Compute $lim_{ntoinfty}frac{tfrac{n}{1}+tfrac{n-1}{2}+dots+tfrac{2}{n-1}+tfrac{1}{n}}{ln(n!)}$ [closed]
$begingroup$
How can I compute the following limit?
$$
lim_{ntoinfty}frac{dfrac{n}{1}+dfrac{n-1}{2}+dots+dfrac{2}{n-1}+dfrac{1}{n}}{ln(n!)}
$$
I have tried lots of methods, I can't get the answer.
Although I think the limit is $0$, I don't know how to explain it. Please, if someone could help me it would be fantastic.
limits
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
$endgroup$
closed as off-topic by RRL, Holo, Namaste, Abcd, Zacky Dec 29 '18 at 12:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, Namaste, Abcd, Zacky
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How can I compute the following limit?
$$
lim_{ntoinfty}frac{dfrac{n}{1}+dfrac{n-1}{2}+dots+dfrac{2}{n-1}+dfrac{1}{n}}{ln(n!)}
$$
I have tried lots of methods, I can't get the answer.
Although I think the limit is $0$, I don't know how to explain it. Please, if someone could help me it would be fantastic.
limits
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
$endgroup$
closed as off-topic by RRL, Holo, Namaste, Abcd, Zacky Dec 29 '18 at 12:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, Namaste, Abcd, Zacky
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:04
3
$begingroup$
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_napprox ln n$ for big n, so this gives us $nln n+ln n-n$ over $ln(n!)$. $ln(n!)$ is about $nln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:07
2
$begingroup$
OK, Wolfram Alpha confirms the limit is $1$.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:10
$begingroup$
Hint: Use Stirling's approximation.
$endgroup$
– Breakfastisready
Dec 28 '18 at 21:21
$begingroup$
Thank you everybody!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
add a comment |
$begingroup$
How can I compute the following limit?
$$
lim_{ntoinfty}frac{dfrac{n}{1}+dfrac{n-1}{2}+dots+dfrac{2}{n-1}+dfrac{1}{n}}{ln(n!)}
$$
I have tried lots of methods, I can't get the answer.
Although I think the limit is $0$, I don't know how to explain it. Please, if someone could help me it would be fantastic.
limits
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
$endgroup$
How can I compute the following limit?
$$
lim_{ntoinfty}frac{dfrac{n}{1}+dfrac{n-1}{2}+dots+dfrac{2}{n-1}+dfrac{1}{n}}{ln(n!)}
$$
I have tried lots of methods, I can't get the answer.
Although I think the limit is $0$, I don't know how to explain it. Please, if someone could help me it would be fantastic.
limits
limits
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
edited Dec 28 '18 at 22:15
Euler Pythagoras
54212
54212
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
asked Dec 28 '18 at 20:57
M.MartinezM.Martinez
343
343
closed as off-topic by RRL, Holo, Namaste, Abcd, Zacky Dec 29 '18 at 12:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, Namaste, Abcd, Zacky
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Holo, Namaste, Abcd, Zacky Dec 29 '18 at 12:41
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Holo, Namaste, Abcd, Zacky
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:04
3
$begingroup$
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_napprox ln n$ for big n, so this gives us $nln n+ln n-n$ over $ln(n!)$. $ln(n!)$ is about $nln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:07
2
$begingroup$
OK, Wolfram Alpha confirms the limit is $1$.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:10
$begingroup$
Hint: Use Stirling's approximation.
$endgroup$
– Breakfastisready
Dec 28 '18 at 21:21
$begingroup$
Thank you everybody!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
add a comment |
3
$begingroup$
Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:04
3
$begingroup$
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_napprox ln n$ for big n, so this gives us $nln n+ln n-n$ over $ln(n!)$. $ln(n!)$ is about $nln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:07
2
$begingroup$
OK, Wolfram Alpha confirms the limit is $1$.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:10
$begingroup$
Hint: Use Stirling's approximation.
$endgroup$
– Breakfastisready
Dec 28 '18 at 21:21
$begingroup$
Thank you everybody!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
3
3
$begingroup$
Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:04
$begingroup$
Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:04
3
3
$begingroup$
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_napprox ln n$ for big n, so this gives us $nln n+ln n-n$ over $ln(n!)$. $ln(n!)$ is about $nln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:07
$begingroup$
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_napprox ln n$ for big n, so this gives us $nln n+ln n-n$ over $ln(n!)$. $ln(n!)$ is about $nln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:07
2
2
$begingroup$
OK, Wolfram Alpha confirms the limit is $1$.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:10
$begingroup$
OK, Wolfram Alpha confirms the limit is $1$.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:10
$begingroup$
Hint: Use Stirling's approximation.
$endgroup$
– Breakfastisready
Dec 28 '18 at 21:21
$begingroup$
Hint: Use Stirling's approximation.
$endgroup$
– Breakfastisready
Dec 28 '18 at 21:21
$begingroup$
Thank you everybody!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
$begingroup$
Thank you everybody!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $H_n:= sumlimits_{k=1}^n frac{1}{k}$.
$ln(n!)sim n(ln(n)-1)$ by Stirling's formula.
As for the numerator, it is $nH_n-frac{1}{2}-frac{2}{3}-cdots-frac{n-1}{n}= nH_n-(n-1)+H_n-1$.
There are very strong estimations for $H_n$, for example $H_n= ln(n)+gamma+O(frac{1}{n})$.
Putting all this together yields that the limit is $1$, you can even obtain a nice error term that the sequence is in fact $1+O(frac{1}{ln(n)})$.
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
1
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
add a comment |
$begingroup$
We can apply Stolz-Cesaro
taking $a_n=(n+1)H_n-n$ and $b_n=ln(n!)$:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty} frac{(n+2)H_{n+1}-(n+1)-(n+1)H_n +n}{ln((n+1)!)-ln(n!)}=lim_{nto infty} frac{H_{n+1}}{ln(n+1)}$$
And since $H_n approx ln n+gamma +Oleft(frac1nright)$ we easily get:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=1Rightarrow lim_{ntoinfty} frac{a_n}{b_n}=1$$
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
$endgroup$
1
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
1
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
1
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $H_n:= sumlimits_{k=1}^n frac{1}{k}$.
$ln(n!)sim n(ln(n)-1)$ by Stirling's formula.
As for the numerator, it is $nH_n-frac{1}{2}-frac{2}{3}-cdots-frac{n-1}{n}= nH_n-(n-1)+H_n-1$.
There are very strong estimations for $H_n$, for example $H_n= ln(n)+gamma+O(frac{1}{n})$.
Putting all this together yields that the limit is $1$, you can even obtain a nice error term that the sequence is in fact $1+O(frac{1}{ln(n)})$.
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
1
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
add a comment |
$begingroup$
Let $H_n:= sumlimits_{k=1}^n frac{1}{k}$.
$ln(n!)sim n(ln(n)-1)$ by Stirling's formula.
As for the numerator, it is $nH_n-frac{1}{2}-frac{2}{3}-cdots-frac{n-1}{n}= nH_n-(n-1)+H_n-1$.
There are very strong estimations for $H_n$, for example $H_n= ln(n)+gamma+O(frac{1}{n})$.
Putting all this together yields that the limit is $1$, you can even obtain a nice error term that the sequence is in fact $1+O(frac{1}{ln(n)})$.
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
$endgroup$
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
1
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
add a comment |
$begingroup$
Let $H_n:= sumlimits_{k=1}^n frac{1}{k}$.
$ln(n!)sim n(ln(n)-1)$ by Stirling's formula.
As for the numerator, it is $nH_n-frac{1}{2}-frac{2}{3}-cdots-frac{n-1}{n}= nH_n-(n-1)+H_n-1$.
There are very strong estimations for $H_n$, for example $H_n= ln(n)+gamma+O(frac{1}{n})$.
Putting all this together yields that the limit is $1$, you can even obtain a nice error term that the sequence is in fact $1+O(frac{1}{ln(n)})$.
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
$endgroup$
Let $H_n:= sumlimits_{k=1}^n frac{1}{k}$.
$ln(n!)sim n(ln(n)-1)$ by Stirling's formula.
As for the numerator, it is $nH_n-frac{1}{2}-frac{2}{3}-cdots-frac{n-1}{n}= nH_n-(n-1)+H_n-1$.
There are very strong estimations for $H_n$, for example $H_n= ln(n)+gamma+O(frac{1}{n})$.
Putting all this together yields that the limit is $1$, you can even obtain a nice error term that the sequence is in fact $1+O(frac{1}{ln(n)})$.
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
answered Dec 28 '18 at 21:41
A. PongráczA. Pongrácz
6,0821929
6,0821929
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
1
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
add a comment |
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
1
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
$begingroup$
Thank you fo helping!!!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28
1
1
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
$begingroup$
I am glad I could help. If you think this answers your question, you should accept the answer.
$endgroup$
– A. Pongrácz
Dec 29 '18 at 7:45
add a comment |
$begingroup$
We can apply Stolz-Cesaro
taking $a_n=(n+1)H_n-n$ and $b_n=ln(n!)$:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty} frac{(n+2)H_{n+1}-(n+1)-(n+1)H_n +n}{ln((n+1)!)-ln(n!)}=lim_{nto infty} frac{H_{n+1}}{ln(n+1)}$$
And since $H_n approx ln n+gamma +Oleft(frac1nright)$ we easily get:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=1Rightarrow lim_{ntoinfty} frac{a_n}{b_n}=1$$
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
$endgroup$
1
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
1
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
1
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
add a comment |
$begingroup$
We can apply Stolz-Cesaro
taking $a_n=(n+1)H_n-n$ and $b_n=ln(n!)$:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty} frac{(n+2)H_{n+1}-(n+1)-(n+1)H_n +n}{ln((n+1)!)-ln(n!)}=lim_{nto infty} frac{H_{n+1}}{ln(n+1)}$$
And since $H_n approx ln n+gamma +Oleft(frac1nright)$ we easily get:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=1Rightarrow lim_{ntoinfty} frac{a_n}{b_n}=1$$
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
$endgroup$
1
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
1
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
1
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
add a comment |
$begingroup$
We can apply Stolz-Cesaro
taking $a_n=(n+1)H_n-n$ and $b_n=ln(n!)$:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty} frac{(n+2)H_{n+1}-(n+1)-(n+1)H_n +n}{ln((n+1)!)-ln(n!)}=lim_{nto infty} frac{H_{n+1}}{ln(n+1)}$$
And since $H_n approx ln n+gamma +Oleft(frac1nright)$ we easily get:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=1Rightarrow lim_{ntoinfty} frac{a_n}{b_n}=1$$
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
$endgroup$
We can apply Stolz-Cesaro
taking $a_n=(n+1)H_n-n$ and $b_n=ln(n!)$:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=lim_{ntoinfty} frac{(n+2)H_{n+1}-(n+1)-(n+1)H_n +n}{ln((n+1)!)-ln(n!)}=lim_{nto infty} frac{H_{n+1}}{ln(n+1)}$$
And since $H_n approx ln n+gamma +Oleft(frac1nright)$ we easily get:
$$l=lim_{nto infty} frac{a_{n+1}-a_n}{b_{n+1}-b_n}=1Rightarrow lim_{ntoinfty} frac{a_n}{b_n}=1$$
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
edited Dec 28 '18 at 22:13
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
answered Dec 28 '18 at 21:56
ZackyZacky
7,87511062
7,87511062
1
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
1
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
1
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
add a comment |
1
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
1
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
1
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
1
1
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
$begingroup$
You have a few typos to fix, like $frac{(n+2)H_{n+1}...}{...}$
$endgroup$
– rtybase
Dec 28 '18 at 22:03
1
1
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
$begingroup$
One more, after all the simplifications, you should have $limlimits_{nrightarrowinfty}frac{H_{n+1}}{ln{(n+1)}}$
$endgroup$
– rtybase
Dec 28 '18 at 22:11
1
1
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
$begingroup$
Thank you for correcting me!
$endgroup$
– Zacky
Dec 28 '18 at 22:14
add a comment |
3
$begingroup$
Can you take a screenshot of what you tried and upload it to us? We like to see what your efforts have been so that we can better help you with this problem.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:04
3
$begingroup$
For what it's worth, Wolfram Alpha says the sum on the top is $(n+1)H_n-n$. $H_napprox ln n$ for big n, so this gives us $nln n+ln n-n$ over $ln(n!)$. $ln(n!)$ is about $nln n$, so I'm guessing the limit is $1$. However, this is very rough reasoning, so don't trust me on that.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:07
2
$begingroup$
OK, Wolfram Alpha confirms the limit is $1$.
$endgroup$
– Noble Mushtak
Dec 28 '18 at 21:10
$begingroup$
Hint: Use Stirling's approximation.
$endgroup$
– Breakfastisready
Dec 28 '18 at 21:21
$begingroup$
Thank you everybody!
$endgroup$
– M.Martinez
Dec 29 '18 at 0:28