Correctness of argument when computing $lim_n n^{frac{1}{n^2}}$
$begingroup$
Say I need to compute $lim_{n to infty} sqrt[n]{sqrt[n]{n}}$ and write
$$lim_{n to infty} sqrt[n]{sqrt[n]{n}} = lim_{n to infty} sqrt[n]{1} = 1$$
by the fact that $lim_{n to infty} sqrt[n]{n} = 1$ (i.e., assume that this fact fact is something that can be used without proof).
Question: how bad (or good) is this solution?
calculus sequences-and-series
asked Jan 21 at 17:29
lumpylumpy
1087
$endgroup$
add a comment |
$begingroup$
Say I need to compute $lim_{n to infty} sqrt[n]{sqrt[n]{n}}$ and write
$$lim_{n to infty} sqrt[n]{sqrt[n]{n}} = lim_{n to infty} sqrt[n]{1} = 1$$
by the fact that $lim_{n to infty} sqrt[n]{n} = 1$ (i.e., assume that this fact fact is something that can be used without proof).
Question: how bad (or good) is this solution?
calculus sequences-and-series
asked Jan 21 at 17:29
lumpylumpy
1087
$endgroup$
1
$begingroup$
That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=sqrt{(n^2)^{1/n^2}} rightarrow 1$ because of the limit that you may use.
$endgroup$
– Mindlack
Jan 21 at 17:33
add a comment |
$begingroup$
Say I need to compute $lim_{n to infty} sqrt[n]{sqrt[n]{n}}$ and write
$$lim_{n to infty} sqrt[n]{sqrt[n]{n}} = lim_{n to infty} sqrt[n]{1} = 1$$
by the fact that $lim_{n to infty} sqrt[n]{n} = 1$ (i.e., assume that this fact fact is something that can be used without proof).
Question: how bad (or good) is this solution?
calculus sequences-and-series
asked Jan 21 at 17:29
lumpylumpy
1087
$endgroup$
Say I need to compute $lim_{n to infty} sqrt[n]{sqrt[n]{n}}$ and write
$$lim_{n to infty} sqrt[n]{sqrt[n]{n}} = lim_{n to infty} sqrt[n]{1} = 1$$
by the fact that $lim_{n to infty} sqrt[n]{n} = 1$ (i.e., assume that this fact fact is something that can be used without proof).
Question: how bad (or good) is this solution?
calculus sequences-and-series
calculus sequences-and-series
asked Jan 21 at 17:29
lumpylumpy
1087
asked Jan 21 at 17:29
lumpylumpy
1087
asked Jan 21 at 17:29
lumpylumpy
1087
asked Jan 21 at 17:29
lumpylumpy
1087
asked Jan 21 at 17:29
lumpylumpy
1087
1087
1
$begingroup$
That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=sqrt{(n^2)^{1/n^2}} rightarrow 1$ because of the limit that you may use.
$endgroup$
– Mindlack
Jan 21 at 17:33
add a comment |
1
$begingroup$
That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=sqrt{(n^2)^{1/n^2}} rightarrow 1$ because of the limit that you may use.
$endgroup$
– Mindlack
Jan 21 at 17:33
1
1
$begingroup$
That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=sqrt{(n^2)^{1/n^2}} rightarrow 1$ because of the limit that you may use.
$endgroup$
– Mindlack
Jan 21 at 17:33
$begingroup$
That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=sqrt{(n^2)^{1/n^2}} rightarrow 1$ because of the limit that you may use.
$endgroup$
– Mindlack
Jan 21 at 17:33
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The solution is not ok that way.
Your argument would fail, for example, for $left( 1+frac{1}{n}right)^n$.
But you can still use the fact $lim_{n to infty} sqrt[n]{n} = 1$ as follows:
$$1 leq sqrt[n]{n} <2 mbox{ for } n mbox{ large enough }$$
$$Rightarrow 1 leq sqrt[n]{sqrt[n]{n}} < sqrt[n]{2} stackrel{n to infty}{longrightarrow}1$$
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
$endgroup$
add a comment |
$begingroup$
$$n^{frac{1}{n^2}} = e^{frac{1}{n^2}ln n}$$
Then use l'Hopital:
$$lim_{nrightarrowinfty}{n^{frac{1}{n^2}} } = e^{lim_{nrightarrowinfty}{frac{ln n}{n^2}}} = e^{lim_{nrightarrowinfty}{frac{1}{2n^2}}} = 1$$
Note that I am allowed to move the limit inside the exponent because of the continuity.
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
Bad: you replace part of the expression with its limit while leaving the rest untouched, which is strictly forbidden.
The faster way consists in calculating the limit of the log, using its asymptotic properties:
$$logBigl(n^{tfrac1{n^2}}Bigr)=frac{log n}{n^2}to 0enspacetext{ as }ntoinfty,$$
so by continuity of the inverse function,
$$ lim_{ntoinfty}n^{tfrac1{n^2}} =mathrm e^0=1.$$
The same proof is valid for any exponent $alpha>0$:
$$ lim_{ntoinfty}n^{tfrac1{n^alpha}}=mathrm e^0=1.$$
answered Jan 21 at 18:07
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
$n>1;$
$1 lt n^{1/n^2} lt (n^2)^{1/n^2}.$
Take the limit.
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
Let $x$ be real and non negative such that $n^{frac{1}{n^2}}=1+x$. Then $n=(1+x)^{n^2}$, so by Bernoulli's inequality we have $nge1+n^2x$ or $xlefrac{1}{n}-frac{1}{n^2}$. Take the limit and you get that $xle0$.
I leave it to you to show is that the root can't actually be strictly smaller than $1$.
answered Jan 21 at 19:26
user495573user495573
1214
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The solution is not ok that way.
Your argument would fail, for example, for $left( 1+frac{1}{n}right)^n$.
But you can still use the fact $lim_{n to infty} sqrt[n]{n} = 1$ as follows:
$$1 leq sqrt[n]{n} <2 mbox{ for } n mbox{ large enough }$$
$$Rightarrow 1 leq sqrt[n]{sqrt[n]{n}} < sqrt[n]{2} stackrel{n to infty}{longrightarrow}1$$
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
$endgroup$
add a comment |
$begingroup$
The solution is not ok that way.
Your argument would fail, for example, for $left( 1+frac{1}{n}right)^n$.
But you can still use the fact $lim_{n to infty} sqrt[n]{n} = 1$ as follows:
$$1 leq sqrt[n]{n} <2 mbox{ for } n mbox{ large enough }$$
$$Rightarrow 1 leq sqrt[n]{sqrt[n]{n}} < sqrt[n]{2} stackrel{n to infty}{longrightarrow}1$$
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
$endgroup$
add a comment |
$begingroup$
The solution is not ok that way.
Your argument would fail, for example, for $left( 1+frac{1}{n}right)^n$.
But you can still use the fact $lim_{n to infty} sqrt[n]{n} = 1$ as follows:
$$1 leq sqrt[n]{n} <2 mbox{ for } n mbox{ large enough }$$
$$Rightarrow 1 leq sqrt[n]{sqrt[n]{n}} < sqrt[n]{2} stackrel{n to infty}{longrightarrow}1$$
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
$endgroup$
The solution is not ok that way.
Your argument would fail, for example, for $left( 1+frac{1}{n}right)^n$.
But you can still use the fact $lim_{n to infty} sqrt[n]{n} = 1$ as follows:
$$1 leq sqrt[n]{n} <2 mbox{ for } n mbox{ large enough }$$
$$Rightarrow 1 leq sqrt[n]{sqrt[n]{n}} < sqrt[n]{2} stackrel{n to infty}{longrightarrow}1$$
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
answered Jan 21 at 17:51
trancelocationtrancelocation
12.3k1826
12.3k1826
add a comment |
add a comment |
$begingroup$
$$n^{frac{1}{n^2}} = e^{frac{1}{n^2}ln n}$$
Then use l'Hopital:
$$lim_{nrightarrowinfty}{n^{frac{1}{n^2}} } = e^{lim_{nrightarrowinfty}{frac{ln n}{n^2}}} = e^{lim_{nrightarrowinfty}{frac{1}{2n^2}}} = 1$$
Note that I am allowed to move the limit inside the exponent because of the continuity.
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
$$n^{frac{1}{n^2}} = e^{frac{1}{n^2}ln n}$$
Then use l'Hopital:
$$lim_{nrightarrowinfty}{n^{frac{1}{n^2}} } = e^{lim_{nrightarrowinfty}{frac{ln n}{n^2}}} = e^{lim_{nrightarrowinfty}{frac{1}{2n^2}}} = 1$$
Note that I am allowed to move the limit inside the exponent because of the continuity.
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
$endgroup$
add a comment |
$begingroup$
$$n^{frac{1}{n^2}} = e^{frac{1}{n^2}ln n}$$
Then use l'Hopital:
$$lim_{nrightarrowinfty}{n^{frac{1}{n^2}} } = e^{lim_{nrightarrowinfty}{frac{ln n}{n^2}}} = e^{lim_{nrightarrowinfty}{frac{1}{2n^2}}} = 1$$
Note that I am allowed to move the limit inside the exponent because of the continuity.
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
$endgroup$
$$n^{frac{1}{n^2}} = e^{frac{1}{n^2}ln n}$$
Then use l'Hopital:
$$lim_{nrightarrowinfty}{n^{frac{1}{n^2}} } = e^{lim_{nrightarrowinfty}{frac{ln n}{n^2}}} = e^{lim_{nrightarrowinfty}{frac{1}{2n^2}}} = 1$$
Note that I am allowed to move the limit inside the exponent because of the continuity.
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
answered Jan 21 at 17:35
lightxbulblightxbulb
945311
945311
add a comment |
add a comment |
$begingroup$
Bad: you replace part of the expression with its limit while leaving the rest untouched, which is strictly forbidden.
The faster way consists in calculating the limit of the log, using its asymptotic properties:
$$logBigl(n^{tfrac1{n^2}}Bigr)=frac{log n}{n^2}to 0enspacetext{ as }ntoinfty,$$
so by continuity of the inverse function,
$$ lim_{ntoinfty}n^{tfrac1{n^2}} =mathrm e^0=1.$$
The same proof is valid for any exponent $alpha>0$:
$$ lim_{ntoinfty}n^{tfrac1{n^alpha}}=mathrm e^0=1.$$
answered Jan 21 at 18:07
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Bad: you replace part of the expression with its limit while leaving the rest untouched, which is strictly forbidden.
The faster way consists in calculating the limit of the log, using its asymptotic properties:
$$logBigl(n^{tfrac1{n^2}}Bigr)=frac{log n}{n^2}to 0enspacetext{ as }ntoinfty,$$
so by continuity of the inverse function,
$$ lim_{ntoinfty}n^{tfrac1{n^2}} =mathrm e^0=1.$$
The same proof is valid for any exponent $alpha>0$:
$$ lim_{ntoinfty}n^{tfrac1{n^alpha}}=mathrm e^0=1.$$
answered Jan 21 at 18:07
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Bad: you replace part of the expression with its limit while leaving the rest untouched, which is strictly forbidden.
The faster way consists in calculating the limit of the log, using its asymptotic properties:
$$logBigl(n^{tfrac1{n^2}}Bigr)=frac{log n}{n^2}to 0enspacetext{ as }ntoinfty,$$
so by continuity of the inverse function,
$$ lim_{ntoinfty}n^{tfrac1{n^2}} =mathrm e^0=1.$$
The same proof is valid for any exponent $alpha>0$:
$$ lim_{ntoinfty}n^{tfrac1{n^alpha}}=mathrm e^0=1.$$
answered Jan 21 at 18:07
BernardBernard
121k740116
$endgroup$
Bad: you replace part of the expression with its limit while leaving the rest untouched, which is strictly forbidden.
The faster way consists in calculating the limit of the log, using its asymptotic properties:
$$logBigl(n^{tfrac1{n^2}}Bigr)=frac{log n}{n^2}to 0enspacetext{ as }ntoinfty,$$
so by continuity of the inverse function,
$$ lim_{ntoinfty}n^{tfrac1{n^2}} =mathrm e^0=1.$$
The same proof is valid for any exponent $alpha>0$:
$$ lim_{ntoinfty}n^{tfrac1{n^alpha}}=mathrm e^0=1.$$
answered Jan 21 at 18:07
BernardBernard
121k740116
answered Jan 21 at 18:07
BernardBernard
121k740116
answered Jan 21 at 18:07
BernardBernard
121k740116
answered Jan 21 at 18:07
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
$n>1;$
$1 lt n^{1/n^2} lt (n^2)^{1/n^2}.$
Take the limit.
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$n>1;$
$1 lt n^{1/n^2} lt (n^2)^{1/n^2}.$
Take the limit.
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$n>1;$
$1 lt n^{1/n^2} lt (n^2)^{1/n^2}.$
Take the limit.
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$n>1;$
$1 lt n^{1/n^2} lt (n^2)^{1/n^2}.$
Take the limit.
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
edited Jan 21 at 18:16
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
answered Jan 21 at 18:11
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
add a comment |
add a comment |
$begingroup$
Let $x$ be real and non negative such that $n^{frac{1}{n^2}}=1+x$. Then $n=(1+x)^{n^2}$, so by Bernoulli's inequality we have $nge1+n^2x$ or $xlefrac{1}{n}-frac{1}{n^2}$. Take the limit and you get that $xle0$.
I leave it to you to show is that the root can't actually be strictly smaller than $1$.
answered Jan 21 at 19:26
user495573user495573
1214
$endgroup$
add a comment |
$begingroup$
Let $x$ be real and non negative such that $n^{frac{1}{n^2}}=1+x$. Then $n=(1+x)^{n^2}$, so by Bernoulli's inequality we have $nge1+n^2x$ or $xlefrac{1}{n}-frac{1}{n^2}$. Take the limit and you get that $xle0$.
I leave it to you to show is that the root can't actually be strictly smaller than $1$.
answered Jan 21 at 19:26
user495573user495573
1214
$endgroup$
add a comment |
$begingroup$
Let $x$ be real and non negative such that $n^{frac{1}{n^2}}=1+x$. Then $n=(1+x)^{n^2}$, so by Bernoulli's inequality we have $nge1+n^2x$ or $xlefrac{1}{n}-frac{1}{n^2}$. Take the limit and you get that $xle0$.
I leave it to you to show is that the root can't actually be strictly smaller than $1$.
answered Jan 21 at 19:26
user495573user495573
1214
$endgroup$
Let $x$ be real and non negative such that $n^{frac{1}{n^2}}=1+x$. Then $n=(1+x)^{n^2}$, so by Bernoulli's inequality we have $nge1+n^2x$ or $xlefrac{1}{n}-frac{1}{n^2}$. Take the limit and you get that $xle0$.
I leave it to you to show is that the root can't actually be strictly smaller than $1$.
answered Jan 21 at 19:26
user495573user495573
1214
answered Jan 21 at 19:26
user495573user495573
1214
answered Jan 21 at 19:26
user495573user495573
1214
answered Jan 21 at 19:26
user495573user495573
1214
1214
add a comment |
add a comment |
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$begingroup$
That is bad, because you did not prove the sequence had a limit and even if you had proved it, you did not prove your limit equality (namely, why can you take back one $n$-th root?). To do it right, you could say $n^{1/n^2}=sqrt{(n^2)^{1/n^2}} rightarrow 1$ because of the limit that you may use.
$endgroup$
– Mindlack
Jan 21 at 17:33