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On the wikipedia page for Gamma function I saw an interesting formula
$$
lim_{nto infty} frac{Gamma(n+alpha)}{Gamma(n)n^alpha} = 1
$$
for all $alphainBbb C$. I couldn't find the source of this and searching here in MSE didn't provide the result I want.

Could anyone show me how this formula is derived?

I'm very inexperienced with properties/identities of $Gamma$ so forgive me if this question is trivial.

The most usual derivation of this would involve the Stirling-Laplace asymptotic for $Gamma(s)$. I'm mildly surprised that this wasn't explicitly worked out in Wiki, or some other easily accessible places on-line.

In fact, a much simpler approach obtains (a stronger version of) this asymptotic via "Watson's Lemma", which is itself easy to completely prove from simple things, going back over 100 years. In various places in the literature, the lemma is in fact called "the oft-reproven Watson's lemma". :)

The case you mention is a simple corollary of the very first example I wrote out in some notes on asymptotic expansions: http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/02d_asymptotics_of_integrals.pdf

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Stirling Asymptotic:
$ds{N! sim root{2pi}, N^{N + 1/2}expo{-N}}$ as
$ds{verts{N} to infty}$.

begin{align}
left.lim_{n to infty}{Gammapars{n + alpha} over Gammapars{n}n^{alpha}},rightvert_{ alpha in mathbb{C}}
& =
lim_{n to infty}{pars{n + alpha - 1}! over
pars{n - 1}!, n^{alpha}}
\[5mm] & =
lim_{n to infty}{root{2pi}pars{n + alpha - 1}^{n + alpha - 1/2}expo{-pars{n + alpha - 1}} over
bracks{root{2pi}pars{n - 1}^{n - 1/2}expo{-pars{n - 1}}}, n^{alpha}}
\[5mm] & =
lim_{n to infty}{n^{n + alpha - 1/2},
bracks{1 + pars{alpha - 1}/n}^{n + alpha - 1/2},expo{-alpha} over
bracks{n^{n - 1/2}pars{1 - 1/n}^{n - 1/2}}, n^{alpha}}
\[5mm] & =
expo{-alpha}lim_{n to infty}
{bracks{1 + pars{alpha - 1}/n}^{n} over pars{1 - 1/n}^{n}}
\[5mm] & =
expo{-alpha},{expo{alpha - 1} over expo{-1}} = bbx{1}
end{align}

Real $boldsymbol{alpha}$

The log-convexity of the Gamma function is shown in this answer.

Suppose that $0lealphale kinmathbb{Z}$, then using the recurrence relation for $Gamma$,
$$
begin{align}
Gamma(n+alpha)
&leGamma(n)^{1-alpha/k},Gamma(n+k)^{alpha/k}\
&leGamma(n)^{1-alpha/k}left(Gamma(n),(n+k)^kright)^{alpha/k}\
&=Gamma(n),(n+k)^alphatag1
end{align}
$$
and
$$
begin{align}
Gamma(n)
&leGamma(n+alpha-k)^{alpha/k}Gamma(n+alpha)^{1-alpha/k}\[6pt]
&leleft(frac{Gamma(n+alpha)}{(n+alpha-k)^k}right)^{alpha/k}Gamma(n+alpha)^{1-alpha/k}\
&=frac{Gamma(n+alpha)}{(n+alpha-k)^alpha}tag2
end{align}
$$
Then we have
$$
left(frac{n+alpha-k}{n}right)^alpha
lefrac{Gamma(n+alpha)}{Gamma(n),n^alpha}
leleft(frac{n+k}{n}right)^alphatag3
$$
and by the Squeeze Theorem, for $alphage0$,
$$
lim_{ntoinfty}frac{Gamma(n+alpha)}{Gamma(n),n^alpha}=1tag4
$$
Substituting $nmapsto n-alpha$ in $(3)$, multiplying by $left(frac{n-alpha}nright)^alpha$, and taking reciprocals, we get
$$
left(frac{n-k}{n}right)^{-alpha}
gefrac{Gamma(n-alpha)}{Gamma(n),n^{-alpha}}
geleft(frac{n-alpha+k}{n}right)^{-alpha}tag5
$$
and by the Squeeze Theorem, for $alphage0$,
$$
lim_{ntoinfty}frac{Gamma(n-alpha)}{Gamma(n),n^{-alpha}}=1tag6
$$

Complex $boldsymbol{alpha}$

Unfortunately, I have not found a way to make the log-convexity argument that works for $alphainmathbb{R}$ work for $alphain mathbb{C}$. About the best I can see, is to use Stirling's Approximation.
$$
Gamma(n)simsqrt{frac{2pi}n}frac{n^n}{e^n}tag7
$$
Applying $(7)$ to $Gamma(n+alpha)$ and $Gamma(n)$, we get
$$
frac{Gamma(n+alpha)}{Gamma(n),n^alpha}simsqrt{frac{n}{n+alpha}}frac{left(1+fracalpha{n}right)^{n+alpha}}{e^alpha}tag8
$$
which yields
$$
lim_{ntoinfty}frac{Gamma(n+alpha)}{Gamma(n),n^alpha}=1tag9
$$

As said, Stirling approximation is the key.

Considering
$$y=frac{Gamma(n+a)}{Gamma(n),n^a}implies log(y)=log (Gamma (a+n))-log (Gamma (n))-a log(n)$$ use Stirling approximation
$$log (Gamma (p))=p (log (p)-1)+frac{1}{2} left(log (2 pi
)-log left({p}right)right)+frac{1}{12 p}-frac{1}{360 p^3}+Oleft(frac{1}{p^5}right)$$ Just apply it and continue with Taylor series to get
$$log(y)=frac{(a-1) a}{2 n}-frac{(a-1) a (2 a-1)}{12 n^2}+frac{(a-1)^2 a^2}{12
n^3}+Oleft(frac{1}{n^4}right)$$ Continue with Taylor
$$y=e^{log(y)}=1+frac{(a-1) a}{2 n}+frac{(a-2) (a-1) a (3 a-1)}{24 n^2}+Oleft(frac{1}{n^3}right)$$

Using it for $n=100$ and $a=5+7i$ the "exact value" would be $(0.798463+ 0.143902 ,i)$ while the above approximation would give $frac{31947}{40000}+frac{1057 }{7500}iapprox (0.798675 +0.140933 i)$.