A series about $n!$ and Riemann zeta function
$begingroup$
Compute
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}.
$$
By the software Mathematica, I find
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}=-frac{2}{3}-frac{zeta left( 1/2 right)}{sqrt{2pi}}.
$$
calculus real-analysis sequences-and-series analysis
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
$endgroup$
add a comment |
$begingroup$
Compute
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}.
$$
By the software Mathematica, I find
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}=-frac{2}{3}-frac{zeta left( 1/2 right)}{sqrt{2pi}}.
$$
calculus real-analysis sequences-and-series analysis
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
$endgroup$
2
$begingroup$
@TheSimpliFire The terms of the series converge to $0$, but there is no reason why the sum has to be close to $0$.
$endgroup$
– Julián Aguirre
Apr 8 '18 at 13:04
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In terms of the Lambert function we have $sum_{ngeq 1}frac{n^{n-1}(-1)^{n+1}}{n!}z^n=W_0(z)$ by Lagrange inversion formula and $frac{1}{sqrt{n}}=int_{0}^{+infty}frac{e^{-ns} ds}{sqrt{pi s}}$ holds by the inverse Laplace transform. In particular $$ t = sum_{ngeq 1}frac{n^{n-1} t^n}{n! e^{nt}} $$ holds for any $t$ in a neighbourhood of the origin, together with $$ 1 = sum_{ngeq 1}frac{n^{n}}{n!e^{nt}}(t^{n-1}-t^n).$$
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– Jack D'Aurizio
Apr 8 '18 at 18:52
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Now it should not be difficult to recover the mentioned identity by playing with the Laplace transform / Ramanujan's master theorem.
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
@MarcoCantarini: on the other hand such coefficients are related to Bernoulli numbers and values of the $zeta$ function, so that can be seen as a convolution between two similar generating functions, related to $logGamma$. This can be useful: $$ frac{n^n}{n!e^n}=frac{1}{2pi}int_{0}^{2pi}left(e^{e^{itheta}-1-itheta}right)^n,dtheta.$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 22:26
add a comment |
$begingroup$
Compute
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}.
$$
By the software Mathematica, I find
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}=-frac{2}{3}-frac{zeta left( 1/2 right)}{sqrt{2pi}}.
$$
calculus real-analysis sequences-and-series analysis
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
$endgroup$
Compute
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}.
$$
By the software Mathematica, I find
$$
sum_{n=1}^{infty}{left( frac{n^n}{n!e^n}-frac{1}{sqrt{2pi n}} right)}=-frac{2}{3}-frac{zeta left( 1/2 right)}{sqrt{2pi}}.
$$
calculus real-analysis sequences-and-series analysis
calculus real-analysis sequences-and-series analysis
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
asked Apr 8 '18 at 10:55
EufiskyEufisky
1,173517
1,173517
2
$begingroup$
@TheSimpliFire The terms of the series converge to $0$, but there is no reason why the sum has to be close to $0$.
$endgroup$
– Julián Aguirre
Apr 8 '18 at 13:04
$begingroup$
In terms of the Lambert function we have $sum_{ngeq 1}frac{n^{n-1}(-1)^{n+1}}{n!}z^n=W_0(z)$ by Lagrange inversion formula and $frac{1}{sqrt{n}}=int_{0}^{+infty}frac{e^{-ns} ds}{sqrt{pi s}}$ holds by the inverse Laplace transform. In particular $$ t = sum_{ngeq 1}frac{n^{n-1} t^n}{n! e^{nt}} $$ holds for any $t$ in a neighbourhood of the origin, together with $$ 1 = sum_{ngeq 1}frac{n^{n}}{n!e^{nt}}(t^{n-1}-t^n).$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
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Now it should not be difficult to recover the mentioned identity by playing with the Laplace transform / Ramanujan's master theorem.
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
@MarcoCantarini: on the other hand such coefficients are related to Bernoulli numbers and values of the $zeta$ function, so that can be seen as a convolution between two similar generating functions, related to $logGamma$. This can be useful: $$ frac{n^n}{n!e^n}=frac{1}{2pi}int_{0}^{2pi}left(e^{e^{itheta}-1-itheta}right)^n,dtheta.$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 22:26
add a comment |
2
$begingroup$
@TheSimpliFire The terms of the series converge to $0$, but there is no reason why the sum has to be close to $0$.
$endgroup$
– Julián Aguirre
Apr 8 '18 at 13:04
$begingroup$
In terms of the Lambert function we have $sum_{ngeq 1}frac{n^{n-1}(-1)^{n+1}}{n!}z^n=W_0(z)$ by Lagrange inversion formula and $frac{1}{sqrt{n}}=int_{0}^{+infty}frac{e^{-ns} ds}{sqrt{pi s}}$ holds by the inverse Laplace transform. In particular $$ t = sum_{ngeq 1}frac{n^{n-1} t^n}{n! e^{nt}} $$ holds for any $t$ in a neighbourhood of the origin, together with $$ 1 = sum_{ngeq 1}frac{n^{n}}{n!e^{nt}}(t^{n-1}-t^n).$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
Now it should not be difficult to recover the mentioned identity by playing with the Laplace transform / Ramanujan's master theorem.
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
@MarcoCantarini: on the other hand such coefficients are related to Bernoulli numbers and values of the $zeta$ function, so that can be seen as a convolution between two similar generating functions, related to $logGamma$. This can be useful: $$ frac{n^n}{n!e^n}=frac{1}{2pi}int_{0}^{2pi}left(e^{e^{itheta}-1-itheta}right)^n,dtheta.$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 22:26
2
2
$begingroup$
@TheSimpliFire The terms of the series converge to $0$, but there is no reason why the sum has to be close to $0$.
$endgroup$
– Julián Aguirre
Apr 8 '18 at 13:04
$begingroup$
@TheSimpliFire The terms of the series converge to $0$, but there is no reason why the sum has to be close to $0$.
$endgroup$
– Julián Aguirre
Apr 8 '18 at 13:04
$begingroup$
In terms of the Lambert function we have $sum_{ngeq 1}frac{n^{n-1}(-1)^{n+1}}{n!}z^n=W_0(z)$ by Lagrange inversion formula and $frac{1}{sqrt{n}}=int_{0}^{+infty}frac{e^{-ns} ds}{sqrt{pi s}}$ holds by the inverse Laplace transform. In particular $$ t = sum_{ngeq 1}frac{n^{n-1} t^n}{n! e^{nt}} $$ holds for any $t$ in a neighbourhood of the origin, together with $$ 1 = sum_{ngeq 1}frac{n^{n}}{n!e^{nt}}(t^{n-1}-t^n).$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
In terms of the Lambert function we have $sum_{ngeq 1}frac{n^{n-1}(-1)^{n+1}}{n!}z^n=W_0(z)$ by Lagrange inversion formula and $frac{1}{sqrt{n}}=int_{0}^{+infty}frac{e^{-ns} ds}{sqrt{pi s}}$ holds by the inverse Laplace transform. In particular $$ t = sum_{ngeq 1}frac{n^{n-1} t^n}{n! e^{nt}} $$ holds for any $t$ in a neighbourhood of the origin, together with $$ 1 = sum_{ngeq 1}frac{n^{n}}{n!e^{nt}}(t^{n-1}-t^n).$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
Now it should not be difficult to recover the mentioned identity by playing with the Laplace transform / Ramanujan's master theorem.
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
Now it should not be difficult to recover the mentioned identity by playing with the Laplace transform / Ramanujan's master theorem.
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
@MarcoCantarini: on the other hand such coefficients are related to Bernoulli numbers and values of the $zeta$ function, so that can be seen as a convolution between two similar generating functions, related to $logGamma$. This can be useful: $$ frac{n^n}{n!e^n}=frac{1}{2pi}int_{0}^{2pi}left(e^{e^{itheta}-1-itheta}right)^n,dtheta.$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 22:26
$begingroup$
@MarcoCantarini: on the other hand such coefficients are related to Bernoulli numbers and values of the $zeta$ function, so that can be seen as a convolution between two similar generating functions, related to $logGamma$. This can be useful: $$ frac{n^n}{n!e^n}=frac{1}{2pi}int_{0}^{2pi}left(e^{e^{itheta}-1-itheta}right)^n,dtheta.$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 22:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Well, $-frac{1}{sqrt{2pi}}zetaleft(tfrac{1}{2}right)$ is the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{1}{sqrt{2pi n}}$, hence the problem boils down to finding the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{n^n}{n!e^n}$. As pointed out in the comments,
$$ W(x) = sum_{ngeq 1}frac{n^{n-1}(-1)^{n-1}}{n!}x^n $$
holds for any $xinleft(-frac{1}{e},frac{1}{e}right)$ by Lagrange inversion theorem, hence
$$ ze^{-z} W'(-ze^{-z})=sum_{ngeq 1}frac{n^n}{n!e^{nz}}z^{n}=frac{z}{1-z} =sum_{ngeq 1}z^ntag{1}$$
holds for any $zin(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.
Similarly, over the same interval
$$ -W(-z e^{-z})=sum_{ngeq 1}frac{n^{n-1}}{n!e^{nz}}z^n = z tag{2}$$
$$ 1=sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n-1}-sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n}=frac{1}{1-z}-frac{z}{1-z}.tag{3}$$
Since $zeta(0)=-frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $zeta$-regularization of $sum_{ngeq 1}frac{n^n}{n!e^n}$ equals $-frac{2}{3}$ as wanted, for instance by computing $sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}$ for any $kinmathbb{N}$:
$$ sum_{ngeq 1}frac{n^{n-2}}{n!e^n}=int_{-1/e}^{1}frac{W(x)}{x},dx = frac{1}{2},qquad sum_{ngeq 1}frac{n^{n-3}}{n!e^n}=-int_{-1/e}^{1}frac{W(x)}{x}(1+log(-x)),dx=frac{5}{12} $$
$$ sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{7}{18},qquad sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{1631}{4320},$$
$$ sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}= frac{1}{Gamma(k)}int_{0}^{1}(1-x)(x-1-log x)^{k-1},dx.tag{4} $$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $zeta$ function complete the proof.
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
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– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
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– Jack D'Aurizio
Apr 10 '18 at 15:13
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Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
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– i707107
Apr 11 '18 at 18:43
2
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
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– i707107
Apr 11 '18 at 19:06
1
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
|
show 1 more comment
$begingroup$
Taking $$Fleft(xright)=sum_{ngeq1}frac{n^{n-1}}{n!e^{n}}x^{n}-frac{1}{sqrt{2pi}}sum_{ngeq1}frac{x^{n}}{n^{3/2}}=-Wleft(-frac{x}{e}right)-frac{mathrm{Li}_{3/2}left(xright)}{sqrt{2pi}},,left|xright|<1$$ where $Wleft(xright)$ is the Lambert $W$ function and $mathrm{Li}_{3/2}left(xright)$ is the Polylogarithm function, we obtain, differentiating both sides,that $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)x^{n-1}=-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}$$ so $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)=lim_{xrightarrow1^{-}}left(-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}right).$$ Now, we know that $$mathrm{Li}_{v}left(zright)=left(Gammaleft(1-vright)left(1-zright)^{v-1}+zetaleft(vright)right)left(1+Oleft(left|1-zright|right)right),vneq1,,zrightarrow1$$ and now we claim $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1}{sqrt{2left(1-xright)}}-frac{2}{3}$$ as $xrightarrow1^{-}$. This is true because, since $$Wleft(zright)sim-1+sqrt{2ze+2}-frac{2}{3}eleft(z+frac{1}{e}right)$$ as $zrightarrow-1/e$, we have $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1-sqrt{2left(1-xright)}+frac{2}{3}left(1-xright)}{xsqrt{2left(1-xright)}-frac{2}{3}left(1-xright)x}$$ $$=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}left(frac{1}{1-sqrt{2-2x}/3}right)right)=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}sum_{kgeq0}left(frac{sqrt{2-2x}}{3}right)^{k}right)$$ $$=frac{1}{x}left(-frac{2}{3}+frac{1}{sqrt{2left(1-xright)}}+Oleft(sqrt{1-x}right)right)$$ then the claim.
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
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2
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
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– Jack D'Aurizio
Apr 9 '18 at 12:34
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A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
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– Jack D'Aurizio
Apr 9 '18 at 12:51
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@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
add a comment |
$begingroup$
This is a general answer to the followup question by Jack D'Aurizio.
Proposition
Let $ninmathbb{N}$. We have the asymptotic expansion
$$
n!sim sqrt{2pi n} frac{n^n}{e^n} left[ 1+ frac1{12n} +frac1{288n^2}-frac{139}{51840n^3}-frac{157}{2488320n^4}+cdots right]
$$
Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $mathbb{Q}[[T]]$.
$$
S(T)=1+ frac1{12}T+frac1{288}T^2-frac{139}{51840}T^3-frac{157}{2488320}T^4 + cdots.
$$
Consider the multiplicative inverse of $S(T)$ in $mathbb{Q}[[T]]$.
$$
S^{-1}(T)=1-frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + cdots.
$$
Let $Y_s(T)=sum_{n=0}^{infty} h_n(s) T^n inmathbb{Q}[s][[T]]$ be defined by
$$
left(frac12 T^2right)^{s-1}sum_{n=0}^{infty} h_n(s) T^n = left[ frac12 T^2 + frac13 T^3 + frac14 T^4+cdots right]^{s-1}.
$$
Then we have
Theorem
$$sum_{n=1}^{infty} n^pleft[ frac{n^n}{n!e^n}- frac1{sqrt{2pi n}} sum_{k=0}^p frac{g_k}{n^k}right]=(-2)^p p!h_{2p+1}(-p) - frac1{sqrt{2pi}}sum_{k=0}^p g_k zetaleft(k+frac12-pright).$$
With $p=0$, it is the original series
$$
sum_{n=1}^{infty} left[frac{n^n}{n!e^n} - frac1{sqrt{2pi n}}right]=-frac23 - frac{zetaleft(frac12right)}{sqrt{2pi}}
$$
With $p=1$, it gives the value of
$$
sum_{n=1}^{infty} left[ frac{n^{n+1}}{n!e^n} - sqrt{frac{n}{2pi}} + frac{1}{12sqrt{2pi n}}right] = -frac 4{135} - frac{zetaleft(-frac12right)}{sqrt{2pi}} + frac{zetaleft(frac12right)}{12sqrt{2pi}}.
$$
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
$endgroup$
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
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– Marco Cantarini
Apr 14 '18 at 7:00
add a comment |
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3 Answers
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$begingroup$
Well, $-frac{1}{sqrt{2pi}}zetaleft(tfrac{1}{2}right)$ is the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{1}{sqrt{2pi n}}$, hence the problem boils down to finding the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{n^n}{n!e^n}$. As pointed out in the comments,
$$ W(x) = sum_{ngeq 1}frac{n^{n-1}(-1)^{n-1}}{n!}x^n $$
holds for any $xinleft(-frac{1}{e},frac{1}{e}right)$ by Lagrange inversion theorem, hence
$$ ze^{-z} W'(-ze^{-z})=sum_{ngeq 1}frac{n^n}{n!e^{nz}}z^{n}=frac{z}{1-z} =sum_{ngeq 1}z^ntag{1}$$
holds for any $zin(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.
Similarly, over the same interval
$$ -W(-z e^{-z})=sum_{ngeq 1}frac{n^{n-1}}{n!e^{nz}}z^n = z tag{2}$$
$$ 1=sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n-1}-sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n}=frac{1}{1-z}-frac{z}{1-z}.tag{3}$$
Since $zeta(0)=-frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $zeta$-regularization of $sum_{ngeq 1}frac{n^n}{n!e^n}$ equals $-frac{2}{3}$ as wanted, for instance by computing $sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}$ for any $kinmathbb{N}$:
$$ sum_{ngeq 1}frac{n^{n-2}}{n!e^n}=int_{-1/e}^{1}frac{W(x)}{x},dx = frac{1}{2},qquad sum_{ngeq 1}frac{n^{n-3}}{n!e^n}=-int_{-1/e}^{1}frac{W(x)}{x}(1+log(-x)),dx=frac{5}{12} $$
$$ sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{7}{18},qquad sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{1631}{4320},$$
$$ sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}= frac{1}{Gamma(k)}int_{0}^{1}(1-x)(x-1-log x)^{k-1},dx.tag{4} $$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $zeta$ function complete the proof.
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
$endgroup$
– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
$endgroup$
– Jack D'Aurizio
Apr 10 '18 at 15:13
$begingroup$
Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
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– i707107
Apr 11 '18 at 18:43
2
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
$endgroup$
– i707107
Apr 11 '18 at 19:06
1
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
|
show 1 more comment
$begingroup$
Well, $-frac{1}{sqrt{2pi}}zetaleft(tfrac{1}{2}right)$ is the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{1}{sqrt{2pi n}}$, hence the problem boils down to finding the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{n^n}{n!e^n}$. As pointed out in the comments,
$$ W(x) = sum_{ngeq 1}frac{n^{n-1}(-1)^{n-1}}{n!}x^n $$
holds for any $xinleft(-frac{1}{e},frac{1}{e}right)$ by Lagrange inversion theorem, hence
$$ ze^{-z} W'(-ze^{-z})=sum_{ngeq 1}frac{n^n}{n!e^{nz}}z^{n}=frac{z}{1-z} =sum_{ngeq 1}z^ntag{1}$$
holds for any $zin(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.
Similarly, over the same interval
$$ -W(-z e^{-z})=sum_{ngeq 1}frac{n^{n-1}}{n!e^{nz}}z^n = z tag{2}$$
$$ 1=sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n-1}-sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n}=frac{1}{1-z}-frac{z}{1-z}.tag{3}$$
Since $zeta(0)=-frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $zeta$-regularization of $sum_{ngeq 1}frac{n^n}{n!e^n}$ equals $-frac{2}{3}$ as wanted, for instance by computing $sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}$ for any $kinmathbb{N}$:
$$ sum_{ngeq 1}frac{n^{n-2}}{n!e^n}=int_{-1/e}^{1}frac{W(x)}{x},dx = frac{1}{2},qquad sum_{ngeq 1}frac{n^{n-3}}{n!e^n}=-int_{-1/e}^{1}frac{W(x)}{x}(1+log(-x)),dx=frac{5}{12} $$
$$ sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{7}{18},qquad sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{1631}{4320},$$
$$ sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}= frac{1}{Gamma(k)}int_{0}^{1}(1-x)(x-1-log x)^{k-1},dx.tag{4} $$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $zeta$ function complete the proof.
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
$endgroup$
– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
$endgroup$
– Jack D'Aurizio
Apr 10 '18 at 15:13
$begingroup$
Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
$endgroup$
– i707107
Apr 11 '18 at 18:43
2
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
$endgroup$
– i707107
Apr 11 '18 at 19:06
1
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
|
show 1 more comment
$begingroup$
Well, $-frac{1}{sqrt{2pi}}zetaleft(tfrac{1}{2}right)$ is the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{1}{sqrt{2pi n}}$, hence the problem boils down to finding the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{n^n}{n!e^n}$. As pointed out in the comments,
$$ W(x) = sum_{ngeq 1}frac{n^{n-1}(-1)^{n-1}}{n!}x^n $$
holds for any $xinleft(-frac{1}{e},frac{1}{e}right)$ by Lagrange inversion theorem, hence
$$ ze^{-z} W'(-ze^{-z})=sum_{ngeq 1}frac{n^n}{n!e^{nz}}z^{n}=frac{z}{1-z} =sum_{ngeq 1}z^ntag{1}$$
holds for any $zin(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.
Similarly, over the same interval
$$ -W(-z e^{-z})=sum_{ngeq 1}frac{n^{n-1}}{n!e^{nz}}z^n = z tag{2}$$
$$ 1=sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n-1}-sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n}=frac{1}{1-z}-frac{z}{1-z}.tag{3}$$
Since $zeta(0)=-frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $zeta$-regularization of $sum_{ngeq 1}frac{n^n}{n!e^n}$ equals $-frac{2}{3}$ as wanted, for instance by computing $sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}$ for any $kinmathbb{N}$:
$$ sum_{ngeq 1}frac{n^{n-2}}{n!e^n}=int_{-1/e}^{1}frac{W(x)}{x},dx = frac{1}{2},qquad sum_{ngeq 1}frac{n^{n-3}}{n!e^n}=-int_{-1/e}^{1}frac{W(x)}{x}(1+log(-x)),dx=frac{5}{12} $$
$$ sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{7}{18},qquad sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{1631}{4320},$$
$$ sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}= frac{1}{Gamma(k)}int_{0}^{1}(1-x)(x-1-log x)^{k-1},dx.tag{4} $$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $zeta$ function complete the proof.
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
$endgroup$
Well, $-frac{1}{sqrt{2pi}}zetaleft(tfrac{1}{2}right)$ is the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{1}{sqrt{2pi n}}$, hence the problem boils down to finding the $zeta$-regularization of the divergent series $sum_{ngeq 1}frac{n^n}{n!e^n}$. As pointed out in the comments,
$$ W(x) = sum_{ngeq 1}frac{n^{n-1}(-1)^{n-1}}{n!}x^n $$
holds for any $xinleft(-frac{1}{e},frac{1}{e}right)$ by Lagrange inversion theorem, hence
$$ ze^{-z} W'(-ze^{-z})=sum_{ngeq 1}frac{n^n}{n!e^{nz}}z^{n}=frac{z}{1-z} =sum_{ngeq 1}z^ntag{1}$$
holds for any $zin(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.
Similarly, over the same interval
$$ -W(-z e^{-z})=sum_{ngeq 1}frac{n^{n-1}}{n!e^{nz}}z^n = z tag{2}$$
$$ 1=sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n-1}-sum_{ngeq 1}frac{n^{n}}{n!e^{nz}}z^{n}=frac{1}{1-z}-frac{z}{1-z}.tag{3}$$
Since $zeta(0)=-frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $zeta$-regularization of $sum_{ngeq 1}frac{n^n}{n!e^n}$ equals $-frac{2}{3}$ as wanted, for instance by computing $sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}$ for any $kinmathbb{N}$:
$$ sum_{ngeq 1}frac{n^{n-2}}{n!e^n}=int_{-1/e}^{1}frac{W(x)}{x},dx = frac{1}{2},qquad sum_{ngeq 1}frac{n^{n-3}}{n!e^n}=-int_{-1/e}^{1}frac{W(x)}{x}(1+log(-x)),dx=frac{5}{12} $$
$$ sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{7}{18},qquad sum_{ngeq 1}frac{n^{n-4}}{n!e^n}=frac{1631}{4320},$$
$$ sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}= frac{1}{Gamma(k)}int_{0}^{1}(1-x)(x-1-log x)^{k-1},dx.tag{4} $$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $zeta$ function complete the proof.
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
edited Apr 9 '18 at 2:41
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
answered Apr 9 '18 at 1:41
Jack D'AurizioJack D'Aurizio
288k33280660
288k33280660
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
$endgroup$
– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
$endgroup$
– Jack D'Aurizio
Apr 10 '18 at 15:13
$begingroup$
Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
$endgroup$
– i707107
Apr 11 '18 at 18:43
2
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
$endgroup$
– i707107
Apr 11 '18 at 19:06
1
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
|
show 1 more comment
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
$endgroup$
– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
$endgroup$
– Jack D'Aurizio
Apr 10 '18 at 15:13
$begingroup$
Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
$endgroup$
– i707107
Apr 11 '18 at 18:43
2
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
$endgroup$
– i707107
Apr 11 '18 at 19:06
1
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
$endgroup$
– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of (4) requires $k$ to be natural number, but we need $k=-1$ for the value of the regularization. Then can you please clarify that step?
$endgroup$
– i707107
Apr 10 '18 at 14:51
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
$endgroup$
– Jack D'Aurizio
Apr 10 '18 at 15:13
$begingroup$
The RHS of $(4)$ requires $k$ to be a real number $>0$. On the other hand $(4)$ leads to an integral representation for $$sum_{ngeq 1}frac{n^{n-1-k}}{n!e^n}-frac{1}{n^{k+1}sqrt{2pi n}}$$ which can be analytically continued by integration by parts.
$endgroup$
– Jack D'Aurizio
Apr 10 '18 at 15:13
$begingroup$
Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
$endgroup$
– i707107
Apr 11 '18 at 18:43
$begingroup$
Thank you. I see that your method can generalize to the series $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$
$endgroup$
– i707107
Apr 11 '18 at 18:43
2
2
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
$endgroup$
– i707107
Apr 11 '18 at 19:06
$begingroup$
Just a little bit simpler formula than your (4), can be obtained either by integration by parts, or using $W'(ze^{-z})$, is $$sum_{ngeq 1} frac{n^n}{n!e^n n^s} = frac1{Gamma(s)} int_0^1 (x-1-log x)^{s-1} dx$$, which is valid at $Re(s)>frac12$.
$endgroup$
– i707107
Apr 11 '18 at 19:06
1
1
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
$begingroup$
We have $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)=-frac 4{135}-frac{zeta(-1/2)}{sqrt{2pi}} + frac{zeta(1/2)}{12sqrt{2pi}}.$$
$endgroup$
– i707107
Apr 12 '18 at 1:39
|
show 1 more comment
$begingroup$
Taking $$Fleft(xright)=sum_{ngeq1}frac{n^{n-1}}{n!e^{n}}x^{n}-frac{1}{sqrt{2pi}}sum_{ngeq1}frac{x^{n}}{n^{3/2}}=-Wleft(-frac{x}{e}right)-frac{mathrm{Li}_{3/2}left(xright)}{sqrt{2pi}},,left|xright|<1$$ where $Wleft(xright)$ is the Lambert $W$ function and $mathrm{Li}_{3/2}left(xright)$ is the Polylogarithm function, we obtain, differentiating both sides,that $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)x^{n-1}=-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}$$ so $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)=lim_{xrightarrow1^{-}}left(-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}right).$$ Now, we know that $$mathrm{Li}_{v}left(zright)=left(Gammaleft(1-vright)left(1-zright)^{v-1}+zetaleft(vright)right)left(1+Oleft(left|1-zright|right)right),vneq1,,zrightarrow1$$ and now we claim $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1}{sqrt{2left(1-xright)}}-frac{2}{3}$$ as $xrightarrow1^{-}$. This is true because, since $$Wleft(zright)sim-1+sqrt{2ze+2}-frac{2}{3}eleft(z+frac{1}{e}right)$$ as $zrightarrow-1/e$, we have $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1-sqrt{2left(1-xright)}+frac{2}{3}left(1-xright)}{xsqrt{2left(1-xright)}-frac{2}{3}left(1-xright)x}$$ $$=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}left(frac{1}{1-sqrt{2-2x}/3}right)right)=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}sum_{kgeq0}left(frac{sqrt{2-2x}}{3}right)^{k}right)$$ $$=frac{1}{x}left(-frac{2}{3}+frac{1}{sqrt{2left(1-xright)}}+Oleft(sqrt{1-x}right)right)$$ then the claim.
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
$endgroup$
2
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:34
$begingroup$
A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:51
$begingroup$
@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
add a comment |
$begingroup$
Taking $$Fleft(xright)=sum_{ngeq1}frac{n^{n-1}}{n!e^{n}}x^{n}-frac{1}{sqrt{2pi}}sum_{ngeq1}frac{x^{n}}{n^{3/2}}=-Wleft(-frac{x}{e}right)-frac{mathrm{Li}_{3/2}left(xright)}{sqrt{2pi}},,left|xright|<1$$ where $Wleft(xright)$ is the Lambert $W$ function and $mathrm{Li}_{3/2}left(xright)$ is the Polylogarithm function, we obtain, differentiating both sides,that $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)x^{n-1}=-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}$$ so $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)=lim_{xrightarrow1^{-}}left(-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}right).$$ Now, we know that $$mathrm{Li}_{v}left(zright)=left(Gammaleft(1-vright)left(1-zright)^{v-1}+zetaleft(vright)right)left(1+Oleft(left|1-zright|right)right),vneq1,,zrightarrow1$$ and now we claim $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1}{sqrt{2left(1-xright)}}-frac{2}{3}$$ as $xrightarrow1^{-}$. This is true because, since $$Wleft(zright)sim-1+sqrt{2ze+2}-frac{2}{3}eleft(z+frac{1}{e}right)$$ as $zrightarrow-1/e$, we have $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1-sqrt{2left(1-xright)}+frac{2}{3}left(1-xright)}{xsqrt{2left(1-xright)}-frac{2}{3}left(1-xright)x}$$ $$=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}left(frac{1}{1-sqrt{2-2x}/3}right)right)=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}sum_{kgeq0}left(frac{sqrt{2-2x}}{3}right)^{k}right)$$ $$=frac{1}{x}left(-frac{2}{3}+frac{1}{sqrt{2left(1-xright)}}+Oleft(sqrt{1-x}right)right)$$ then the claim.
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
$endgroup$
2
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:34
$begingroup$
A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:51
$begingroup$
@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
add a comment |
$begingroup$
Taking $$Fleft(xright)=sum_{ngeq1}frac{n^{n-1}}{n!e^{n}}x^{n}-frac{1}{sqrt{2pi}}sum_{ngeq1}frac{x^{n}}{n^{3/2}}=-Wleft(-frac{x}{e}right)-frac{mathrm{Li}_{3/2}left(xright)}{sqrt{2pi}},,left|xright|<1$$ where $Wleft(xright)$ is the Lambert $W$ function and $mathrm{Li}_{3/2}left(xright)$ is the Polylogarithm function, we obtain, differentiating both sides,that $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)x^{n-1}=-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}$$ so $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)=lim_{xrightarrow1^{-}}left(-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}right).$$ Now, we know that $$mathrm{Li}_{v}left(zright)=left(Gammaleft(1-vright)left(1-zright)^{v-1}+zetaleft(vright)right)left(1+Oleft(left|1-zright|right)right),vneq1,,zrightarrow1$$ and now we claim $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1}{sqrt{2left(1-xright)}}-frac{2}{3}$$ as $xrightarrow1^{-}$. This is true because, since $$Wleft(zright)sim-1+sqrt{2ze+2}-frac{2}{3}eleft(z+frac{1}{e}right)$$ as $zrightarrow-1/e$, we have $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1-sqrt{2left(1-xright)}+frac{2}{3}left(1-xright)}{xsqrt{2left(1-xright)}-frac{2}{3}left(1-xright)x}$$ $$=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}left(frac{1}{1-sqrt{2-2x}/3}right)right)=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}sum_{kgeq0}left(frac{sqrt{2-2x}}{3}right)^{k}right)$$ $$=frac{1}{x}left(-frac{2}{3}+frac{1}{sqrt{2left(1-xright)}}+Oleft(sqrt{1-x}right)right)$$ then the claim.
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
$endgroup$
Taking $$Fleft(xright)=sum_{ngeq1}frac{n^{n-1}}{n!e^{n}}x^{n}-frac{1}{sqrt{2pi}}sum_{ngeq1}frac{x^{n}}{n^{3/2}}=-Wleft(-frac{x}{e}right)-frac{mathrm{Li}_{3/2}left(xright)}{sqrt{2pi}},,left|xright|<1$$ where $Wleft(xright)$ is the Lambert $W$ function and $mathrm{Li}_{3/2}left(xright)$ is the Polylogarithm function, we obtain, differentiating both sides,that $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)x^{n-1}=-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}$$ so $$sum_{ngeq1}left(frac{n^{n}}{n!e^{n}}-frac{1}{sqrt{2pi n}}right)=lim_{xrightarrow1^{-}}left(-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}-frac{mathrm{Li}_{1/2}left(xright)}{xsqrt{2pi}}right).$$ Now, we know that $$mathrm{Li}_{v}left(zright)=left(Gammaleft(1-vright)left(1-zright)^{v-1}+zetaleft(vright)right)left(1+Oleft(left|1-zright|right)right),vneq1,,zrightarrow1$$ and now we claim $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1}{sqrt{2left(1-xright)}}-frac{2}{3}$$ as $xrightarrow1^{-}$. This is true because, since $$Wleft(zright)sim-1+sqrt{2ze+2}-frac{2}{3}eleft(z+frac{1}{e}right)$$ as $zrightarrow-1/e$, we have $$-frac{Wleft(-frac{x}{e}right)}{xleft(Wleft(-frac{x}{e}right)+1right)}simfrac{1-sqrt{2left(1-xright)}+frac{2}{3}left(1-xright)}{xsqrt{2left(1-xright)}-frac{2}{3}left(1-xright)x}$$ $$=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}left(frac{1}{1-sqrt{2-2x}/3}right)right)=frac{1}{x}left(-1+frac{1}{sqrt{2left(1-xright)}}sum_{kgeq0}left(frac{sqrt{2-2x}}{3}right)^{k}right)$$ $$=frac{1}{x}left(-frac{2}{3}+frac{1}{sqrt{2left(1-xright)}}+Oleft(sqrt{1-x}right)right)$$ then the claim.
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
edited Apr 17 '18 at 14:50
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
answered Apr 9 '18 at 8:52
Marco CantariniMarco Cantarini
29.1k23372
29.1k23372
2
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:34
$begingroup$
A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:51
$begingroup$
@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
add a comment |
2
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:34
$begingroup$
A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:51
$begingroup$
@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
2
2
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:34
$begingroup$
(+1) This is a nice way of avoiding a double transformation via $mathcal{L},mathcal{L}^{-1}$. Good job, as usual. Is the next step to compute $$sum_{ngeq 1}left(frac{n^{n+1}}{n!e^n}-sqrt{frac{n}{2pi}}+frac{1}{12sqrt{2pi n}}right)$$ ?
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:34
$begingroup$
A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:51
$begingroup$
A minor fix: I believe $sqrt{2ze+1}$ should be $sqrt{2ze+2}$. It does not affect anything, the relevant term of the asymptotics stays $-frac{2}{3}(ez+1)$.
$endgroup$
– Jack D'Aurizio
Apr 9 '18 at 12:51
$begingroup$
@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
$begingroup$
@JackD'Aurizio Thank you, fixed. I upvoted your answer this morning (as usual, I like your answers more than mine). Yes, I think the next step is the series you write. I'm not sure that my method works also in this case, probably there is something to change.
$endgroup$
– Marco Cantarini
Apr 9 '18 at 13:55
add a comment |
$begingroup$
This is a general answer to the followup question by Jack D'Aurizio.
Proposition
Let $ninmathbb{N}$. We have the asymptotic expansion
$$
n!sim sqrt{2pi n} frac{n^n}{e^n} left[ 1+ frac1{12n} +frac1{288n^2}-frac{139}{51840n^3}-frac{157}{2488320n^4}+cdots right]
$$
Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $mathbb{Q}[[T]]$.
$$
S(T)=1+ frac1{12}T+frac1{288}T^2-frac{139}{51840}T^3-frac{157}{2488320}T^4 + cdots.
$$
Consider the multiplicative inverse of $S(T)$ in $mathbb{Q}[[T]]$.
$$
S^{-1}(T)=1-frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + cdots.
$$
Let $Y_s(T)=sum_{n=0}^{infty} h_n(s) T^n inmathbb{Q}[s][[T]]$ be defined by
$$
left(frac12 T^2right)^{s-1}sum_{n=0}^{infty} h_n(s) T^n = left[ frac12 T^2 + frac13 T^3 + frac14 T^4+cdots right]^{s-1}.
$$
Then we have
Theorem
$$sum_{n=1}^{infty} n^pleft[ frac{n^n}{n!e^n}- frac1{sqrt{2pi n}} sum_{k=0}^p frac{g_k}{n^k}right]=(-2)^p p!h_{2p+1}(-p) - frac1{sqrt{2pi}}sum_{k=0}^p g_k zetaleft(k+frac12-pright).$$
With $p=0$, it is the original series
$$
sum_{n=1}^{infty} left[frac{n^n}{n!e^n} - frac1{sqrt{2pi n}}right]=-frac23 - frac{zetaleft(frac12right)}{sqrt{2pi}}
$$
With $p=1$, it gives the value of
$$
sum_{n=1}^{infty} left[ frac{n^{n+1}}{n!e^n} - sqrt{frac{n}{2pi}} + frac{1}{12sqrt{2pi n}}right] = -frac 4{135} - frac{zetaleft(-frac12right)}{sqrt{2pi}} + frac{zetaleft(frac12right)}{12sqrt{2pi}}.
$$
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
$endgroup$
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
$endgroup$
– Marco Cantarini
Apr 14 '18 at 7:00
add a comment |
$begingroup$
This is a general answer to the followup question by Jack D'Aurizio.
Proposition
Let $ninmathbb{N}$. We have the asymptotic expansion
$$
n!sim sqrt{2pi n} frac{n^n}{e^n} left[ 1+ frac1{12n} +frac1{288n^2}-frac{139}{51840n^3}-frac{157}{2488320n^4}+cdots right]
$$
Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $mathbb{Q}[[T]]$.
$$
S(T)=1+ frac1{12}T+frac1{288}T^2-frac{139}{51840}T^3-frac{157}{2488320}T^4 + cdots.
$$
Consider the multiplicative inverse of $S(T)$ in $mathbb{Q}[[T]]$.
$$
S^{-1}(T)=1-frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + cdots.
$$
Let $Y_s(T)=sum_{n=0}^{infty} h_n(s) T^n inmathbb{Q}[s][[T]]$ be defined by
$$
left(frac12 T^2right)^{s-1}sum_{n=0}^{infty} h_n(s) T^n = left[ frac12 T^2 + frac13 T^3 + frac14 T^4+cdots right]^{s-1}.
$$
Then we have
Theorem
$$sum_{n=1}^{infty} n^pleft[ frac{n^n}{n!e^n}- frac1{sqrt{2pi n}} sum_{k=0}^p frac{g_k}{n^k}right]=(-2)^p p!h_{2p+1}(-p) - frac1{sqrt{2pi}}sum_{k=0}^p g_k zetaleft(k+frac12-pright).$$
With $p=0$, it is the original series
$$
sum_{n=1}^{infty} left[frac{n^n}{n!e^n} - frac1{sqrt{2pi n}}right]=-frac23 - frac{zetaleft(frac12right)}{sqrt{2pi}}
$$
With $p=1$, it gives the value of
$$
sum_{n=1}^{infty} left[ frac{n^{n+1}}{n!e^n} - sqrt{frac{n}{2pi}} + frac{1}{12sqrt{2pi n}}right] = -frac 4{135} - frac{zetaleft(-frac12right)}{sqrt{2pi}} + frac{zetaleft(frac12right)}{12sqrt{2pi}}.
$$
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
$endgroup$
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
$endgroup$
– Marco Cantarini
Apr 14 '18 at 7:00
add a comment |
$begingroup$
This is a general answer to the followup question by Jack D'Aurizio.
Proposition
Let $ninmathbb{N}$. We have the asymptotic expansion
$$
n!sim sqrt{2pi n} frac{n^n}{e^n} left[ 1+ frac1{12n} +frac1{288n^2}-frac{139}{51840n^3}-frac{157}{2488320n^4}+cdots right]
$$
Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $mathbb{Q}[[T]]$.
$$
S(T)=1+ frac1{12}T+frac1{288}T^2-frac{139}{51840}T^3-frac{157}{2488320}T^4 + cdots.
$$
Consider the multiplicative inverse of $S(T)$ in $mathbb{Q}[[T]]$.
$$
S^{-1}(T)=1-frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + cdots.
$$
Let $Y_s(T)=sum_{n=0}^{infty} h_n(s) T^n inmathbb{Q}[s][[T]]$ be defined by
$$
left(frac12 T^2right)^{s-1}sum_{n=0}^{infty} h_n(s) T^n = left[ frac12 T^2 + frac13 T^3 + frac14 T^4+cdots right]^{s-1}.
$$
Then we have
Theorem
$$sum_{n=1}^{infty} n^pleft[ frac{n^n}{n!e^n}- frac1{sqrt{2pi n}} sum_{k=0}^p frac{g_k}{n^k}right]=(-2)^p p!h_{2p+1}(-p) - frac1{sqrt{2pi}}sum_{k=0}^p g_k zetaleft(k+frac12-pright).$$
With $p=0$, it is the original series
$$
sum_{n=1}^{infty} left[frac{n^n}{n!e^n} - frac1{sqrt{2pi n}}right]=-frac23 - frac{zetaleft(frac12right)}{sqrt{2pi}}
$$
With $p=1$, it gives the value of
$$
sum_{n=1}^{infty} left[ frac{n^{n+1}}{n!e^n} - sqrt{frac{n}{2pi}} + frac{1}{12sqrt{2pi n}}right] = -frac 4{135} - frac{zetaleft(-frac12right)}{sqrt{2pi}} + frac{zetaleft(frac12right)}{12sqrt{2pi}}.
$$
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
$endgroup$
This is a general answer to the followup question by Jack D'Aurizio.
Proposition
Let $ninmathbb{N}$. We have the asymptotic expansion
$$
n!sim sqrt{2pi n} frac{n^n}{e^n} left[ 1+ frac1{12n} +frac1{288n^2}-frac{139}{51840n^3}-frac{157}{2488320n^4}+cdots right]
$$
Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $mathbb{Q}[[T]]$.
$$
S(T)=1+ frac1{12}T+frac1{288}T^2-frac{139}{51840}T^3-frac{157}{2488320}T^4 + cdots.
$$
Consider the multiplicative inverse of $S(T)$ in $mathbb{Q}[[T]]$.
$$
S^{-1}(T)=1-frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + cdots.
$$
Let $Y_s(T)=sum_{n=0}^{infty} h_n(s) T^n inmathbb{Q}[s][[T]]$ be defined by
$$
left(frac12 T^2right)^{s-1}sum_{n=0}^{infty} h_n(s) T^n = left[ frac12 T^2 + frac13 T^3 + frac14 T^4+cdots right]^{s-1}.
$$
Then we have
Theorem
$$sum_{n=1}^{infty} n^pleft[ frac{n^n}{n!e^n}- frac1{sqrt{2pi n}} sum_{k=0}^p frac{g_k}{n^k}right]=(-2)^p p!h_{2p+1}(-p) - frac1{sqrt{2pi}}sum_{k=0}^p g_k zetaleft(k+frac12-pright).$$
With $p=0$, it is the original series
$$
sum_{n=1}^{infty} left[frac{n^n}{n!e^n} - frac1{sqrt{2pi n}}right]=-frac23 - frac{zetaleft(frac12right)}{sqrt{2pi}}
$$
With $p=1$, it gives the value of
$$
sum_{n=1}^{infty} left[ frac{n^{n+1}}{n!e^n} - sqrt{frac{n}{2pi}} + frac{1}{12sqrt{2pi n}}right] = -frac 4{135} - frac{zetaleft(-frac12right)}{sqrt{2pi}} + frac{zetaleft(frac12right)}{12sqrt{2pi}}.
$$
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
edited Apr 13 '18 at 21:44
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
answered Apr 13 '18 at 21:38
i707107i707107
11.9k21547
11.9k21547
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
$endgroup$
– Marco Cantarini
Apr 14 '18 at 7:00
add a comment |
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
$endgroup$
– Marco Cantarini
Apr 14 '18 at 7:00
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
$endgroup$
– Marco Cantarini
Apr 14 '18 at 7:00
$begingroup$
(+1) Very nice!!! My remark was wrong, thank you to find the right sense! :D
$endgroup$
– Marco Cantarini
Apr 14 '18 at 7:00
add a comment |
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$begingroup$
@TheSimpliFire The terms of the series converge to $0$, but there is no reason why the sum has to be close to $0$.
$endgroup$
– Julián Aguirre
Apr 8 '18 at 13:04
$begingroup$
In terms of the Lambert function we have $sum_{ngeq 1}frac{n^{n-1}(-1)^{n+1}}{n!}z^n=W_0(z)$ by Lagrange inversion formula and $frac{1}{sqrt{n}}=int_{0}^{+infty}frac{e^{-ns} ds}{sqrt{pi s}}$ holds by the inverse Laplace transform. In particular $$ t = sum_{ngeq 1}frac{n^{n-1} t^n}{n! e^{nt}} $$ holds for any $t$ in a neighbourhood of the origin, together with $$ 1 = sum_{ngeq 1}frac{n^{n}}{n!e^{nt}}(t^{n-1}-t^n).$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
Now it should not be difficult to recover the mentioned identity by playing with the Laplace transform / Ramanujan's master theorem.
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 18:52
$begingroup$
@MarcoCantarini: on the other hand such coefficients are related to Bernoulli numbers and values of the $zeta$ function, so that can be seen as a convolution between two similar generating functions, related to $logGamma$. This can be useful: $$ frac{n^n}{n!e^n}=frac{1}{2pi}int_{0}^{2pi}left(e^{e^{itheta}-1-itheta}right)^n,dtheta.$$
$endgroup$
– Jack D'Aurizio
Apr 8 '18 at 22:26