How can I prove $ lim_{nrightarrowinfty}...
How can I prove that the limit
$$
lim_{nrightarrowinfty} left(frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)right)=sqrt{frac{e}{2}}
$$
limits products
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
asked Dec 1 '18 at 14:30
avan1235
1776
closed as off-topic by Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen Dec 1 '18 at 21:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
How can I prove that the limit
$$
lim_{nrightarrowinfty} left(frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)right)=sqrt{frac{e}{2}}
$$
limits products
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
asked Dec 1 '18 at 14:30
avan1235
1776
closed as off-topic by Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen Dec 1 '18 at 21:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
I would try multiplying the numerator and denominator by $2^nn!$ to make the denominator $(2n)!$ and use Stirling's approximation. I'm pretty sure it will work.
– Ross Millikan
Dec 1 '18 at 14:38
The eulers-constant tag is for the Euler–Mascheroni constant $gamma approx 0.577216$, not $e approx 2.71828$.
– Bladewood
Dec 1 '18 at 16:56
add a comment |
How can I prove that the limit
$$
lim_{nrightarrowinfty} left(frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)right)=sqrt{frac{e}{2}}
$$
limits products
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
asked Dec 1 '18 at 14:30
avan1235
1776
How can I prove that the limit
$$
lim_{nrightarrowinfty} left(frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)right)=sqrt{frac{e}{2}}
$$
limits products
limits products
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
asked Dec 1 '18 at 14:30
avan1235
1776
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
asked Dec 1 '18 at 14:30
avan1235
1776
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
edited Dec 1 '18 at 17:59
Paramanand Singh
48.9k555158
48.9k555158
asked Dec 1 '18 at 14:30
avan1235
1776
asked Dec 1 '18 at 14:30
avan1235
1776
asked Dec 1 '18 at 14:30
avan1235
1776
1776
closed as off-topic by Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen Dec 1 '18 at 21:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen Dec 1 '18 at 21:10
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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Paramanand Singh, Lee Mosher, amWhy, José Carlos Santos, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
I would try multiplying the numerator and denominator by $2^nn!$ to make the denominator $(2n)!$ and use Stirling's approximation. I'm pretty sure it will work.
– Ross Millikan
Dec 1 '18 at 14:38
The eulers-constant tag is for the Euler–Mascheroni constant $gamma approx 0.577216$, not $e approx 2.71828$.
– Bladewood
Dec 1 '18 at 16:56
add a comment |
I would try multiplying the numerator and denominator by $2^nn!$ to make the denominator $(2n)!$ and use Stirling's approximation. I'm pretty sure it will work.
– Ross Millikan
Dec 1 '18 at 14:38
The eulers-constant tag is for the Euler–Mascheroni constant $gamma approx 0.577216$, not $e approx 2.71828$.
– Bladewood
Dec 1 '18 at 16:56
I would try multiplying the numerator and denominator by $2^nn!$ to make the denominator $(2n)!$ and use Stirling's approximation. I'm pretty sure it will work.
– Ross Millikan
Dec 1 '18 at 14:38
I would try multiplying the numerator and denominator by $2^nn!$ to make the denominator $(2n)!$ and use Stirling's approximation. I'm pretty sure it will work.
– Ross Millikan
Dec 1 '18 at 14:38
The eulers-constant tag is for the Euler–Mascheroni constant $gamma approx 0.577216$, not $e approx 2.71828$.
– Bladewood
Dec 1 '18 at 16:56
The eulers-constant tag is for the Euler–Mascheroni constant $gamma approx 0.577216$, not $e approx 2.71828$.
– Bladewood
Dec 1 '18 at 16:56
add a comment |
2 Answers
2
active
oldest
votes
HINT
We have that
$$frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{1}{e^n}left(frac{2^nn!(2n+1)^n}{(2n)!}right)$$
then use Stirling's approximation for $n!$.
answered Dec 1 '18 at 14:39
gimusi
1
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
|
show 1 more comment
If you ant to go beyond, consider
$$prod_{k=1}^n (2k-1)=frac{2^n, Gamma left(n+frac{1}{2}right)}{sqrt{pi }}$$
$$a_n=frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{sqrt{pi } (2 e)^{-n} (2 n+1)^n}{Gamma left(n+frac{1}{2}right)}$$ Take logarithms and use Stirling approximation as already said in answers to get
$$log(a_n)=left(frac{1}{2}-frac{log (2)}{2}right)-frac{1}{12 n}+frac{1}{24
n^2}+Oleft(frac{1}{n^3}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=sqrt{frac{e}{2}}left(1-frac{1}{12 n}+frac{13}{288 n^2}+Oleft(frac{1}{n^3}right)right)$$ which gives the limit, how it is approached and an approximation.
Using $n=10$, the exact solution is
$$a_{10}=frac{29417779503}{1154725 e^{10}}approx 1.156609$$ while the above expansion gives
$$a_{10}=frac{28573 sqrt{frac{e}{2}}}{28800}approx 1.156633$$
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
We have that
$$frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{1}{e^n}left(frac{2^nn!(2n+1)^n}{(2n)!}right)$$
then use Stirling's approximation for $n!$.
answered Dec 1 '18 at 14:39
gimusi
1
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
|
show 1 more comment
HINT
We have that
$$frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{1}{e^n}left(frac{2^nn!(2n+1)^n}{(2n)!}right)$$
then use Stirling's approximation for $n!$.
answered Dec 1 '18 at 14:39
gimusi
1
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
|
show 1 more comment
HINT
We have that
$$frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{1}{e^n}left(frac{2^nn!(2n+1)^n}{(2n)!}right)$$
then use Stirling's approximation for $n!$.
answered Dec 1 '18 at 14:39
gimusi
1
HINT
We have that
$$frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{1}{e^n}left(frac{2^nn!(2n+1)^n}{(2n)!}right)$$
then use Stirling's approximation for $n!$.
answered Dec 1 '18 at 14:39
gimusi
1
answered Dec 1 '18 at 14:39
gimusi
1
answered Dec 1 '18 at 14:39
gimusi
1
answered Dec 1 '18 at 14:39
gimusi
1
1
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
|
show 1 more comment
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
We can write $(2n+1)^n=(2n)^n(1+1/2n)^n$ and second factor tends to $sqrt{e} $. Thus the problem is reduced to showing that $dfrac{(4n/e)^nn!}{(2n)!}to 1/sqrt{2}$ and this reminds me of another similar question.
– Paramanand Singh
Dec 1 '18 at 14:46
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
And that question is math.stackexchange.com/q/2847386/72031
– Paramanand Singh
Dec 1 '18 at 14:47
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
@ParamanandSingh That's a nice observation and a very good idea to simplify further!
– gimusi
Dec 1 '18 at 14:48
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
Thus we can avoid Stirling and instead use the powerful but not so famous lemma mentioned in this answer math.stackexchange.com/a/2847768/72031
– Paramanand Singh
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
@ParamanandSingh The first one was already among my fav.
– gimusi
Dec 1 '18 at 14:50
|
show 1 more comment
If you ant to go beyond, consider
$$prod_{k=1}^n (2k-1)=frac{2^n, Gamma left(n+frac{1}{2}right)}{sqrt{pi }}$$
$$a_n=frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{sqrt{pi } (2 e)^{-n} (2 n+1)^n}{Gamma left(n+frac{1}{2}right)}$$ Take logarithms and use Stirling approximation as already said in answers to get
$$log(a_n)=left(frac{1}{2}-frac{log (2)}{2}right)-frac{1}{12 n}+frac{1}{24
n^2}+Oleft(frac{1}{n^3}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=sqrt{frac{e}{2}}left(1-frac{1}{12 n}+frac{13}{288 n^2}+Oleft(frac{1}{n^3}right)right)$$ which gives the limit, how it is approached and an approximation.
Using $n=10$, the exact solution is
$$a_{10}=frac{29417779503}{1154725 e^{10}}approx 1.156609$$ while the above expansion gives
$$a_{10}=frac{28573 sqrt{frac{e}{2}}}{28800}approx 1.156633$$
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
add a comment |
If you ant to go beyond, consider
$$prod_{k=1}^n (2k-1)=frac{2^n, Gamma left(n+frac{1}{2}right)}{sqrt{pi }}$$
$$a_n=frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{sqrt{pi } (2 e)^{-n} (2 n+1)^n}{Gamma left(n+frac{1}{2}right)}$$ Take logarithms and use Stirling approximation as already said in answers to get
$$log(a_n)=left(frac{1}{2}-frac{log (2)}{2}right)-frac{1}{12 n}+frac{1}{24
n^2}+Oleft(frac{1}{n^3}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=sqrt{frac{e}{2}}left(1-frac{1}{12 n}+frac{13}{288 n^2}+Oleft(frac{1}{n^3}right)right)$$ which gives the limit, how it is approached and an approximation.
Using $n=10$, the exact solution is
$$a_{10}=frac{29417779503}{1154725 e^{10}}approx 1.156609$$ while the above expansion gives
$$a_{10}=frac{28573 sqrt{frac{e}{2}}}{28800}approx 1.156633$$
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
add a comment |
If you ant to go beyond, consider
$$prod_{k=1}^n (2k-1)=frac{2^n, Gamma left(n+frac{1}{2}right)}{sqrt{pi }}$$
$$a_n=frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{sqrt{pi } (2 e)^{-n} (2 n+1)^n}{Gamma left(n+frac{1}{2}right)}$$ Take logarithms and use Stirling approximation as already said in answers to get
$$log(a_n)=left(frac{1}{2}-frac{log (2)}{2}right)-frac{1}{12 n}+frac{1}{24
n^2}+Oleft(frac{1}{n^3}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=sqrt{frac{e}{2}}left(1-frac{1}{12 n}+frac{13}{288 n^2}+Oleft(frac{1}{n^3}right)right)$$ which gives the limit, how it is approached and an approximation.
Using $n=10$, the exact solution is
$$a_{10}=frac{29417779503}{1154725 e^{10}}approx 1.156609$$ while the above expansion gives
$$a_{10}=frac{28573 sqrt{frac{e}{2}}}{28800}approx 1.156633$$
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
If you ant to go beyond, consider
$$prod_{k=1}^n (2k-1)=frac{2^n, Gamma left(n+frac{1}{2}right)}{sqrt{pi }}$$
$$a_n=frac{1}{e^n}left(frac{(2n+1)^n}{prod_{k=1}^{n}(2k-1)}right)=frac{sqrt{pi } (2 e)^{-n} (2 n+1)^n}{Gamma left(n+frac{1}{2}right)}$$ Take logarithms and use Stirling approximation as already said in answers to get
$$log(a_n)=left(frac{1}{2}-frac{log (2)}{2}right)-frac{1}{12 n}+frac{1}{24
n^2}+Oleft(frac{1}{n^3}right)$$ Continue with Taylor
$$a_n=e^{log(a_n)}=sqrt{frac{e}{2}}left(1-frac{1}{12 n}+frac{13}{288 n^2}+Oleft(frac{1}{n^3}right)right)$$ which gives the limit, how it is approached and an approximation.
Using $n=10$, the exact solution is
$$a_{10}=frac{29417779503}{1154725 e^{10}}approx 1.156609$$ while the above expansion gives
$$a_{10}=frac{28573 sqrt{frac{e}{2}}}{28800}approx 1.156633$$
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
answered Dec 1 '18 at 16:42
Claude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
I would try multiplying the numerator and denominator by $2^nn!$ to make the denominator $(2n)!$ and use Stirling's approximation. I'm pretty sure it will work.
– Ross Millikan
Dec 1 '18 at 14:38
The eulers-constant tag is for the Euler–Mascheroni constant $gamma approx 0.577216$, not $e approx 2.71828$.
– Bladewood
Dec 1 '18 at 16:56