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Series $sum n!exp(-n^2)$ is convergent?
The Next CEO of Stack OverflowA question about convergent seriesNeed help applying the root test for: $sumlimits_n=1^inftyleft(frac2e^-8n-1right)^n$Applying Stirling's formula in testing for convergence of a sumConvergent series problemAbsolute vs Conditional Convergence SeriesProof convergence of seriesWhy does this series $sum_n=0^infty frac(n!)^2(2n)!$ converge?How to figure out if this series is convergent or divergent?Help with Determining whether this series is convergent or divergent.Convergence of $sumlimits_m= 0^infty frac(-3)^m+2^m3^m+2^m$
$begingroup$
Consider the series $sum n!exp(-n^2)$.
By the ratio test, I've proven that it is convergent (unless I made a horrible mistake).
However, I'm trying to prove this using Stirling's Formula for $n!$.
I can't go any further than:
$ n!exp(-n^2) sim sqrt2pi exp((n + frac12)ln(n) - n - n^2) $
How can I proceed? Or is there any other method without using the ratio test or Stirling's approximation? Or maybe I'm wrong and it's divergent?
Thank you for your answers!
sequences-and-series convergence
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
$endgroup$
add a comment |
$begingroup$
Consider the series $sum n!exp(-n^2)$.
By the ratio test, I've proven that it is convergent (unless I made a horrible mistake).
However, I'm trying to prove this using Stirling's Formula for $n!$.
I can't go any further than:
$ n!exp(-n^2) sim sqrt2pi exp((n + frac12)ln(n) - n - n^2) $
How can I proceed? Or is there any other method without using the ratio test or Stirling's approximation? Or maybe I'm wrong and it's divergent?
Thank you for your answers!
sequences-and-series convergence
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
$endgroup$
$begingroup$
You surely can estimate what of $nlog n$ and $n^2$ grows faster with $n$...
$endgroup$
– user
Mar 19 at 22:36
$begingroup$
You can use Stirling's formula and the root test Sirling's formula alone won't work, as $fsim g$ doesn't imply $mathrm e^f simmathrm e^g$.
$endgroup$
– Bernard
Mar 19 at 23:06
$begingroup$
The ratio test is much easier: you just consider $lim_ntoinftyfracn+1e^2n+1=0$.
$endgroup$
– egreg
Mar 19 at 23:07
add a comment |
$begingroup$
Consider the series $sum n!exp(-n^2)$.
By the ratio test, I've proven that it is convergent (unless I made a horrible mistake).
However, I'm trying to prove this using Stirling's Formula for $n!$.
I can't go any further than:
$ n!exp(-n^2) sim sqrt2pi exp((n + frac12)ln(n) - n - n^2) $
How can I proceed? Or is there any other method without using the ratio test or Stirling's approximation? Or maybe I'm wrong and it's divergent?
Thank you for your answers!
sequences-and-series convergence
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
$endgroup$
Consider the series $sum n!exp(-n^2)$.
By the ratio test, I've proven that it is convergent (unless I made a horrible mistake).
However, I'm trying to prove this using Stirling's Formula for $n!$.
I can't go any further than:
$ n!exp(-n^2) sim sqrt2pi exp((n + frac12)ln(n) - n - n^2) $
How can I proceed? Or is there any other method without using the ratio test or Stirling's approximation? Or maybe I'm wrong and it's divergent?
Thank you for your answers!
sequences-and-series convergence
sequences-and-series convergence
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
asked Mar 19 at 22:24
Mau_Ssy7Mau_Ssy7
488
488
$begingroup$
You surely can estimate what of $nlog n$ and $n^2$ grows faster with $n$...
$endgroup$
– user
Mar 19 at 22:36
$begingroup$
You can use Stirling's formula and the root test Sirling's formula alone won't work, as $fsim g$ doesn't imply $mathrm e^f simmathrm e^g$.
$endgroup$
– Bernard
Mar 19 at 23:06
$begingroup$
The ratio test is much easier: you just consider $lim_ntoinftyfracn+1e^2n+1=0$.
$endgroup$
– egreg
Mar 19 at 23:07
add a comment |
$begingroup$
You surely can estimate what of $nlog n$ and $n^2$ grows faster with $n$...
$endgroup$
– user
Mar 19 at 22:36
$begingroup$
You can use Stirling's formula and the root test Sirling's formula alone won't work, as $fsim g$ doesn't imply $mathrm e^f simmathrm e^g$.
$endgroup$
– Bernard
Mar 19 at 23:06
$begingroup$
The ratio test is much easier: you just consider $lim_ntoinftyfracn+1e^2n+1=0$.
$endgroup$
– egreg
Mar 19 at 23:07
$begingroup$
You surely can estimate what of $nlog n$ and $n^2$ grows faster with $n$...
$endgroup$
– user
Mar 19 at 22:36
$begingroup$
You surely can estimate what of $nlog n$ and $n^2$ grows faster with $n$...
$endgroup$
– user
Mar 19 at 22:36
$begingroup$
You can use Stirling's formula and the root test Sirling's formula alone won't work, as $fsim g$ doesn't imply $mathrm e^f simmathrm e^g$.
$endgroup$
– Bernard
Mar 19 at 23:06
$begingroup$
You can use Stirling's formula and the root test Sirling's formula alone won't work, as $fsim g$ doesn't imply $mathrm e^f simmathrm e^g$.
$endgroup$
– Bernard
Mar 19 at 23:06
$begingroup$
The ratio test is much easier: you just consider $lim_ntoinftyfracn+1e^2n+1=0$.
$endgroup$
– egreg
Mar 19 at 23:07
$begingroup$
The ratio test is much easier: you just consider $lim_ntoinftyfracn+1e^2n+1=0$.
$endgroup$
– egreg
Mar 19 at 23:07
add a comment |
1 Answer
1
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$begingroup$
Your expression from Stirling is good :
$$n! exp(-n^2) sim sqrt2pi expleft(left(n+ frac12right)ln(n) - n^2right)=oleft(expleft(- fracn^22 right) right)$$
and $expleft(- fracn^22 right) $ is the general term of a convergent series. Therefore, your series converges.
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
$endgroup$
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1 Answer
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1 Answer
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$begingroup$
Your expression from Stirling is good :
$$n! exp(-n^2) sim sqrt2pi expleft(left(n+ frac12right)ln(n) - n^2right)=oleft(expleft(- fracn^22 right) right)$$
and $expleft(- fracn^22 right) $ is the general term of a convergent series. Therefore, your series converges.
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
$endgroup$
add a comment |
$begingroup$
Your expression from Stirling is good :
$$n! exp(-n^2) sim sqrt2pi expleft(left(n+ frac12right)ln(n) - n^2right)=oleft(expleft(- fracn^22 right) right)$$
and $expleft(- fracn^22 right) $ is the general term of a convergent series. Therefore, your series converges.
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
$endgroup$
add a comment |
$begingroup$
Your expression from Stirling is good :
$$n! exp(-n^2) sim sqrt2pi expleft(left(n+ frac12right)ln(n) - n^2right)=oleft(expleft(- fracn^22 right) right)$$
and $expleft(- fracn^22 right) $ is the general term of a convergent series. Therefore, your series converges.
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
$endgroup$
Your expression from Stirling is good :
$$n! exp(-n^2) sim sqrt2pi expleft(left(n+ frac12right)ln(n) - n^2right)=oleft(expleft(- fracn^22 right) right)$$
and $expleft(- fracn^22 right) $ is the general term of a convergent series. Therefore, your series converges.
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
answered Mar 19 at 22:29
TheSilverDoeTheSilverDoe
4,947215
4,947215
add a comment |
add a comment |
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$begingroup$
You surely can estimate what of $nlog n$ and $n^2$ grows faster with $n$...
$endgroup$
– user
Mar 19 at 22:36
$begingroup$
You can use Stirling's formula and the root test Sirling's formula alone won't work, as $fsim g$ doesn't imply $mathrm e^f simmathrm e^g$.
$endgroup$
– Bernard
Mar 19 at 23:06
$begingroup$
The ratio test is much easier: you just consider $lim_ntoinftyfracn+1e^2n+1=0$.
$endgroup$
– egreg
Mar 19 at 23:07