Prove that $sum_{n=1}^{infty} big(frac{1}{4}big)^n {2n choose n}$ diverges [duplicate]
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This question already has an answer here:
Show that $sum_{n=1}^{infty} frac{(2n)!}{4^n(n!)^2}$ is divergent
5 answers
I need to prove that the following series $$ sum_{n=1}^{infty} bigg(frac{1}{4}bigg)^n {2n choose n}$$ diverges.
I've managed to change it to the following formula: $$ 4^n=(1+1)^{2n}=bigg(sum_{k=0}^n{nchoose k}bigg)^2 text{ and } {2nchoose n}=sum_{k=0}^n{nchoose k}^2 text{ so }$$ $$sum_{n=1}^infty bigg(frac{1}{4}bigg)^n {2n choose n} =sum_{n=1}^infty frac{sum_{k=0}^n{nchoose k}^2}{left(sum_{k=0}^n{nchoose k}right)^2} $$ Where do I go from here? I know that the limit of series' sequence goes to $0$ but that isn't helpful really since I need to prove that it diverges.
Also all the ratio tests and root test are inconclusive. I already checked that. Think of 4 as variable $a$. I need to check for which $a$ this series converges. I already know that for $a>4$ it does, but I need to prove that for $a=4$ it diverges. There was a mistake in my calculating of limit in Raabe's test which concluded that this series is indeed divergent.
real-analysis sequences-and-series
edited Dec 8 '18 at 13:22
amWhy
1
asked Dec 8 '18 at 11:26
VictorVictor
786
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marked as duplicate by Nosrati, Xander Henderson, Lord Shark the Unknown, user10354138, Cesareo Dec 9 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Show that $sum_{n=1}^{infty} frac{(2n)!}{4^n(n!)^2}$ is divergent
5 answers
I need to prove that the following series $$ sum_{n=1}^{infty} bigg(frac{1}{4}bigg)^n {2n choose n}$$ diverges.
I've managed to change it to the following formula: $$ 4^n=(1+1)^{2n}=bigg(sum_{k=0}^n{nchoose k}bigg)^2 text{ and } {2nchoose n}=sum_{k=0}^n{nchoose k}^2 text{ so }$$ $$sum_{n=1}^infty bigg(frac{1}{4}bigg)^n {2n choose n} =sum_{n=1}^infty frac{sum_{k=0}^n{nchoose k}^2}{left(sum_{k=0}^n{nchoose k}right)^2} $$ Where do I go from here? I know that the limit of series' sequence goes to $0$ but that isn't helpful really since I need to prove that it diverges.
Also all the ratio tests and root test are inconclusive. I already checked that. Think of 4 as variable $a$. I need to check for which $a$ this series converges. I already know that for $a>4$ it does, but I need to prove that for $a=4$ it diverges. There was a mistake in my calculating of limit in Raabe's test which concluded that this series is indeed divergent.
real-analysis sequences-and-series
edited Dec 8 '18 at 13:22
amWhy
1
asked Dec 8 '18 at 11:26
VictorVictor
786
$endgroup$
marked as duplicate by Nosrati, Xander Henderson, Lord Shark the Unknown, user10354138, Cesareo Dec 9 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Try bounding your last term by the RMS inequality.
$endgroup$
– siddharth64
Dec 8 '18 at 11:31
1
$begingroup$
Use either Stirling formula (for a one-liner) or a refinement of the ratio test called the Raabe's test.
$endgroup$
– Did
Dec 8 '18 at 11:42
1
$begingroup$
See here: math.stackexchange.com/q/2857889/515527
$endgroup$
– Zacky
Dec 8 '18 at 12:17
add a comment |
$begingroup$
This question already has an answer here:
Show that $sum_{n=1}^{infty} frac{(2n)!}{4^n(n!)^2}$ is divergent
5 answers
I need to prove that the following series $$ sum_{n=1}^{infty} bigg(frac{1}{4}bigg)^n {2n choose n}$$ diverges.
I've managed to change it to the following formula: $$ 4^n=(1+1)^{2n}=bigg(sum_{k=0}^n{nchoose k}bigg)^2 text{ and } {2nchoose n}=sum_{k=0}^n{nchoose k}^2 text{ so }$$ $$sum_{n=1}^infty bigg(frac{1}{4}bigg)^n {2n choose n} =sum_{n=1}^infty frac{sum_{k=0}^n{nchoose k}^2}{left(sum_{k=0}^n{nchoose k}right)^2} $$ Where do I go from here? I know that the limit of series' sequence goes to $0$ but that isn't helpful really since I need to prove that it diverges.
Also all the ratio tests and root test are inconclusive. I already checked that. Think of 4 as variable $a$. I need to check for which $a$ this series converges. I already know that for $a>4$ it does, but I need to prove that for $a=4$ it diverges. There was a mistake in my calculating of limit in Raabe's test which concluded that this series is indeed divergent.
real-analysis sequences-and-series
edited Dec 8 '18 at 13:22
amWhy
1
asked Dec 8 '18 at 11:26
VictorVictor
786
$endgroup$
This question already has an answer here:
Show that $sum_{n=1}^{infty} frac{(2n)!}{4^n(n!)^2}$ is divergent
5 answers
I need to prove that the following series $$ sum_{n=1}^{infty} bigg(frac{1}{4}bigg)^n {2n choose n}$$ diverges.
I've managed to change it to the following formula: $$ 4^n=(1+1)^{2n}=bigg(sum_{k=0}^n{nchoose k}bigg)^2 text{ and } {2nchoose n}=sum_{k=0}^n{nchoose k}^2 text{ so }$$ $$sum_{n=1}^infty bigg(frac{1}{4}bigg)^n {2n choose n} =sum_{n=1}^infty frac{sum_{k=0}^n{nchoose k}^2}{left(sum_{k=0}^n{nchoose k}right)^2} $$ Where do I go from here? I know that the limit of series' sequence goes to $0$ but that isn't helpful really since I need to prove that it diverges.
Also all the ratio tests and root test are inconclusive. I already checked that. Think of 4 as variable $a$. I need to check for which $a$ this series converges. I already know that for $a>4$ it does, but I need to prove that for $a=4$ it diverges. There was a mistake in my calculating of limit in Raabe's test which concluded that this series is indeed divergent.
This question already has an answer here:
Show that $sum_{n=1}^{infty} frac{(2n)!}{4^n(n!)^2}$ is divergent
5 answers
real-analysis sequences-and-series
real-analysis sequences-and-series
edited Dec 8 '18 at 13:22
amWhy
1
asked Dec 8 '18 at 11:26
VictorVictor
786
edited Dec 8 '18 at 13:22
amWhy
1
asked Dec 8 '18 at 11:26
VictorVictor
786
edited Dec 8 '18 at 13:22
amWhy
1
edited Dec 8 '18 at 13:22
amWhy
1
edited Dec 8 '18 at 13:22
amWhy
1
1
asked Dec 8 '18 at 11:26
VictorVictor
786
asked Dec 8 '18 at 11:26
VictorVictor
786
asked Dec 8 '18 at 11:26
VictorVictor
786
786
marked as duplicate by Nosrati, Xander Henderson, Lord Shark the Unknown, user10354138, Cesareo Dec 9 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Nosrati, Xander Henderson, Lord Shark the Unknown, user10354138, Cesareo Dec 9 '18 at 9:30
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Try bounding your last term by the RMS inequality.
$endgroup$
– siddharth64
Dec 8 '18 at 11:31
1
$begingroup$
Use either Stirling formula (for a one-liner) or a refinement of the ratio test called the Raabe's test.
$endgroup$
– Did
Dec 8 '18 at 11:42
1
$begingroup$
See here: math.stackexchange.com/q/2857889/515527
$endgroup$
– Zacky
Dec 8 '18 at 12:17
add a comment |
$begingroup$
Try bounding your last term by the RMS inequality.
$endgroup$
– siddharth64
Dec 8 '18 at 11:31
1
$begingroup$
Use either Stirling formula (for a one-liner) or a refinement of the ratio test called the Raabe's test.
$endgroup$
– Did
Dec 8 '18 at 11:42
1
$begingroup$
See here: math.stackexchange.com/q/2857889/515527
$endgroup$
– Zacky
Dec 8 '18 at 12:17
$begingroup$
Try bounding your last term by the RMS inequality.
$endgroup$
– siddharth64
Dec 8 '18 at 11:31
$begingroup$
Try bounding your last term by the RMS inequality.
$endgroup$
– siddharth64
Dec 8 '18 at 11:31
1
1
$begingroup$
Use either Stirling formula (for a one-liner) or a refinement of the ratio test called the Raabe's test.
$endgroup$
– Did
Dec 8 '18 at 11:42
$begingroup$
Use either Stirling formula (for a one-liner) or a refinement of the ratio test called the Raabe's test.
$endgroup$
– Did
Dec 8 '18 at 11:42
1
1
$begingroup$
See here: math.stackexchange.com/q/2857889/515527
$endgroup$
– Zacky
Dec 8 '18 at 12:17
$begingroup$
See here: math.stackexchange.com/q/2857889/515527
$endgroup$
– Zacky
Dec 8 '18 at 12:17
add a comment |
2 Answers
2
active
oldest
votes
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Let's calculate $lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)$,where $a_n= bigg(frac{1}{4}bigg)^n {2n choose n}$ and
$a_{n+1}=bigg(frac{1}{4}bigg)^{n+1} {2n+2 choose n+1}$.
begin{align}
lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)&=lim_{nto infty}nleft(4frac{(n+1)^2}{(2n+1)(2n+2)}-1right)\
&=lim_{nto infty}frac{n}{2n+1}\
&=frac{1}{2}
end{align}
Since $frac{1}{2} lt 1$,by Raabe's test, we can conclude that the series diverges.
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
1
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
1
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
add a comment |
$begingroup$
HINT
We have that
$$bigg(frac{1}{4}bigg)^n {2n choose n}sim frac{1}{sqrt{pi n}}$$
then refer to limit comparison test.
Refer to the related
- Elementary central binomial coefficient estimates
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
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@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
Something wrong?
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– gimusi
Dec 8 '18 at 13:09
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's calculate $lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)$,where $a_n= bigg(frac{1}{4}bigg)^n {2n choose n}$ and
$a_{n+1}=bigg(frac{1}{4}bigg)^{n+1} {2n+2 choose n+1}$.
begin{align}
lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)&=lim_{nto infty}nleft(4frac{(n+1)^2}{(2n+1)(2n+2)}-1right)\
&=lim_{nto infty}frac{n}{2n+1}\
&=frac{1}{2}
end{align}
Since $frac{1}{2} lt 1$,by Raabe's test, we can conclude that the series diverges.
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
1
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
1
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
add a comment |
$begingroup$
Let's calculate $lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)$,where $a_n= bigg(frac{1}{4}bigg)^n {2n choose n}$ and
$a_{n+1}=bigg(frac{1}{4}bigg)^{n+1} {2n+2 choose n+1}$.
begin{align}
lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)&=lim_{nto infty}nleft(4frac{(n+1)^2}{(2n+1)(2n+2)}-1right)\
&=lim_{nto infty}frac{n}{2n+1}\
&=frac{1}{2}
end{align}
Since $frac{1}{2} lt 1$,by Raabe's test, we can conclude that the series diverges.
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
1
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
1
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
add a comment |
$begingroup$
Let's calculate $lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)$,where $a_n= bigg(frac{1}{4}bigg)^n {2n choose n}$ and
$a_{n+1}=bigg(frac{1}{4}bigg)^{n+1} {2n+2 choose n+1}$.
begin{align}
lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)&=lim_{nto infty}nleft(4frac{(n+1)^2}{(2n+1)(2n+2)}-1right)\
&=lim_{nto infty}frac{n}{2n+1}\
&=frac{1}{2}
end{align}
Since $frac{1}{2} lt 1$,by Raabe's test, we can conclude that the series diverges.
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
$endgroup$
Let's calculate $lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)$,where $a_n= bigg(frac{1}{4}bigg)^n {2n choose n}$ and
$a_{n+1}=bigg(frac{1}{4}bigg)^{n+1} {2n+2 choose n+1}$.
begin{align}
lim_{nto infty}nleft(frac{a_n}{a_{n+1}}-1right)&=lim_{nto infty}nleft(4frac{(n+1)^2}{(2n+1)(2n+2)}-1right)\
&=lim_{nto infty}frac{n}{2n+1}\
&=frac{1}{2}
end{align}
Since $frac{1}{2} lt 1$,by Raabe's test, we can conclude that the series diverges.
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
edited Dec 8 '18 at 12:31
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
answered Dec 8 '18 at 12:24
Thomas ShelbyThomas Shelby
2,710421
2,710421
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
1
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
1
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
add a comment |
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
1
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
1
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
$begingroup$
I've double checked my calculations. I think Raabe's test works fine here.
$endgroup$
– Thomas Shelby
Dec 8 '18 at 12:26
1
1
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
$begingroup$
Sure it does, and your computation is fine. (+1)
$endgroup$
– Did
Dec 8 '18 at 12:30
1
1
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
$begingroup$
Calculated this limit once again and you were indeed right, there was a mistake and this limit is $frac{1}{2}$
$endgroup$
– Victor
Dec 8 '18 at 12:55
add a comment |
$begingroup$
HINT
We have that
$$bigg(frac{1}{4}bigg)^n {2n choose n}sim frac{1}{sqrt{pi n}}$$
then refer to limit comparison test.
Refer to the related
- Elementary central binomial coefficient estimates
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
$begingroup$
@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
Something wrong?
$endgroup$
– gimusi
Dec 8 '18 at 13:09
add a comment |
$begingroup$
HINT
We have that
$$bigg(frac{1}{4}bigg)^n {2n choose n}sim frac{1}{sqrt{pi n}}$$
then refer to limit comparison test.
Refer to the related
- Elementary central binomial coefficient estimates
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
$endgroup$
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
$begingroup$
@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
Something wrong?
$endgroup$
– gimusi
Dec 8 '18 at 13:09
add a comment |
$begingroup$
HINT
We have that
$$bigg(frac{1}{4}bigg)^n {2n choose n}sim frac{1}{sqrt{pi n}}$$
then refer to limit comparison test.
Refer to the related
- Elementary central binomial coefficient estimates
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
$endgroup$
HINT
We have that
$$bigg(frac{1}{4}bigg)^n {2n choose n}sim frac{1}{sqrt{pi n}}$$
then refer to limit comparison test.
Refer to the related
- Elementary central binomial coefficient estimates
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
edited Dec 8 '18 at 12:04
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
answered Dec 8 '18 at 11:47
gimusigimusi
92.8k84494
92.8k84494
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
$begingroup$
@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
Something wrong?
$endgroup$
– gimusi
Dec 8 '18 at 13:09
add a comment |
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
$begingroup$
@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
Something wrong?
$endgroup$
– gimusi
Dec 8 '18 at 13:09
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
$begingroup$
The root test is also inconclusive.
$endgroup$
– Kemono Chen
Dec 8 '18 at 11:49
$begingroup$
@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
@KemonoChen Ops yes of course! I see it now...Thanks
$endgroup$
– gimusi
Dec 8 '18 at 11:54
$begingroup$
Something wrong?
$endgroup$
– gimusi
Dec 8 '18 at 13:09
$begingroup$
Something wrong?
$endgroup$
– gimusi
Dec 8 '18 at 13:09
add a comment |
$begingroup$
Try bounding your last term by the RMS inequality.
$endgroup$
– siddharth64
Dec 8 '18 at 11:31
1
$begingroup$
Use either Stirling formula (for a one-liner) or a refinement of the ratio test called the Raabe's test.
$endgroup$
– Did
Dec 8 '18 at 11:42
1
$begingroup$
See here: math.stackexchange.com/q/2857889/515527
$endgroup$
– Zacky
Dec 8 '18 at 12:17