Mixing
Infinite Series examples [closed]
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$$ sum _{n=0}^{infty }:left(frac{4^{n:}n!:n!}{left(2nright)!}right)$$ is diverges, I need steps.
Thanks
calculus sequences-and-series
edited Jan 14 at 6:24
mrtaurho
4,66131235
asked Jan 12 at 13:40
DiDoDiDo
42
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closed as off-topic by José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn Jan 28 at 16:52
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$ sum _{n=0}^{infty }:left(frac{4^{n:}n!:n!}{left(2nright)!}right)$$ is diverges, I need steps.
Thanks
calculus sequences-and-series
edited Jan 14 at 6:24
mrtaurho
4,66131235
asked Jan 12 at 13:40
DiDoDiDo
42
$endgroup$
closed as off-topic by José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn Jan 28 at 16:52
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." – José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
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Hint: use Stirling formula for the factorial (you should obtain the Wallis product)
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– Raymond Manzoni
Jan 12 at 13:44
1
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If $a_n$ is as in the question, I think $a_n to infty$, and therefore $sum a_n$ diverges.
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– GEdgar
Jan 12 at 14:28
add a comment |
$begingroup$
$$ sum _{n=0}^{infty }:left(frac{4^{n:}n!:n!}{left(2nright)!}right)$$ is diverges, I need steps.
Thanks
calculus sequences-and-series
edited Jan 14 at 6:24
mrtaurho
4,66131235
asked Jan 12 at 13:40
DiDoDiDo
42
$endgroup$
$$ sum _{n=0}^{infty }:left(frac{4^{n:}n!:n!}{left(2nright)!}right)$$ is diverges, I need steps.
Thanks
calculus sequences-and-series
calculus sequences-and-series
edited Jan 14 at 6:24
mrtaurho
4,66131235
asked Jan 12 at 13:40
DiDoDiDo
42
edited Jan 14 at 6:24
mrtaurho
4,66131235
asked Jan 12 at 13:40
DiDoDiDo
42
edited Jan 14 at 6:24
mrtaurho
4,66131235
edited Jan 14 at 6:24
mrtaurho
4,66131235
edited Jan 14 at 6:24
mrtaurho
4,66131235
4,66131235
asked Jan 12 at 13:40
DiDoDiDo
42
asked Jan 12 at 13:40
DiDoDiDo
42
asked Jan 12 at 13:40
DiDoDiDo
42
42
closed as off-topic by José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn Jan 28 at 16:52
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." – José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn Jan 28 at 16:52
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." – José Carlos Santos, amWhy, RRL, Xander Henderson, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hint: use Stirling formula for the factorial (you should obtain the Wallis product)
$endgroup$
– Raymond Manzoni
Jan 12 at 13:44
1
$begingroup$
If $a_n$ is as in the question, I think $a_n to infty$, and therefore $sum a_n$ diverges.
$endgroup$
– GEdgar
Jan 12 at 14:28
add a comment |
$begingroup$
Hint: use Stirling formula for the factorial (you should obtain the Wallis product)
$endgroup$
– Raymond Manzoni
Jan 12 at 13:44
1
$begingroup$
If $a_n$ is as in the question, I think $a_n to infty$, and therefore $sum a_n$ diverges.
$endgroup$
– GEdgar
Jan 12 at 14:28
$begingroup$
Hint: use Stirling formula for the factorial (you should obtain the Wallis product)
$endgroup$
– Raymond Manzoni
Jan 12 at 13:44
$begingroup$
Hint: use Stirling formula for the factorial (you should obtain the Wallis product)
$endgroup$
– Raymond Manzoni
Jan 12 at 13:44
1
1
$begingroup$
If $a_n$ is as in the question, I think $a_n to infty$, and therefore $sum a_n$ diverges.
$endgroup$
– GEdgar
Jan 12 at 14:28
$begingroup$
If $a_n$ is as in the question, I think $a_n to infty$, and therefore $sum a_n$ diverges.
$endgroup$
– GEdgar
Jan 12 at 14:28
add a comment |
3 Answers
3
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votes
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D'Alembert criterion is inconclusive beacause
$frac{a_{n+1}}{a_n}=frac{frac{4^{n+1};(n+1)!;(n+1)!}{left(2(n+1)right)!}}{frac{4^{n};n!;n!}{left(2nright)!}}=frac{4(n+1)(n+1)}{(2n+1)(2n+2)} rightarrow 1$ as $n rightarrow infty$, but as @Raymond Manzoni said, we could use Raabe's test. We are calculating $nleft(frac{a_n}{a_{n+1}}-1right)$. We have
begin{equation*}
nleft(frac{frac{4^n;n!;n!}{(2n)!}}{frac{4^{n+1};(n+1)!;(n+1)!}{(2n+2)!}}-1right)=nleft(frac{(2n+1)(2n+2)}{4(n+1)^2}-1right)=frac{-n^2-n}{2n^2+4n+4} \ =frac{-n(n+1)}{2(n+1)^2}leq 0,
end{equation*}
for all $n in mathbb{N}$ beacause the numerator is always negative and denominator positive, so it is especially less than $1$.
answered Jan 12 at 13:51
trulekstruleks
2114
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Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
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– DiDo
Jan 12 at 13:57
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(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
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– Raymond Manzoni
Jan 12 at 14:12
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Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
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– truleks
Jan 12 at 14:16
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@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
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– Raymond Manzoni
Jan 12 at 14:19
1
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Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
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– DiDo
Jan 12 at 15:01
|
show 3 more comments
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Consider
$$sum _{n=0}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}=sum _{n=0}^{p }frac{4^{n:}n!:n!}{left(2nright)!}+sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}$$
Now, consider
$$a_n=frac{4^{n:}n!:n!}{left(2nright)!}implies log(a_n)=n log(4)+2log(n!)-log((2n)!)$$ and use Stirling approximation. This would give, for large $n$
$$log(a_n)=frac{1}{2} left(log (pi )+log left({n}right)right)+Oleft(frac{1}{n}right)implies a_nsimeq {sqrt{pi n}}$$ which shows that
$$sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}to infty$$
Edit
In fact, you could easily show, that for any $n geq 1$, $a_n >sqrt{pi n}$ which makes
$$sum _{n=0}^{p}frac{4^{n:}n!:n!}{left(2nright)!}=2+sum _{n=1}^{p }frac{4^{n:}n!:n!}{left(2nright)!} > 2+sum _{n=1}^{p } sqrt{pi n}=2+sqrt{pi } H_p^{left(-frac{1}{2}right)}$$ where appears generalized harmonic numbers. Using the asymptotics
$$ H_p^{left(-frac{1}{2}right)}=frac{2 }{3}p^{3/2}+frac{1}{2}p^{1/2}+zeta
left(-frac{1}{2}right)+Oleft(frac{1}{p^{1/2}}right)$$
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
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Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
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– Raymond Manzoni
Jan 12 at 15:34
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@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
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– Claude Leibovici
Jan 12 at 15:37
add a comment |
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Using Stirling's approximation for factorial and for large enough $n$ we have $${4^ncdot (n!)^2over (2n)!}{sim{4^n left[sqrt{2pi n}left({nover e}right)^nright]^2over {sqrt{4pi n}}left({2nover e}right)^{2n}}\={4^ncdot {2pi n}cdotleft({nover e}right)^{2n}over {sqrt{4pi n}}left({2nover e}right)^{2n}}\= {4^ncdot {2pi n}over {sqrt{4pi n}}cdot {4^n}}\=sqrt{pi n}}$$therefore for $n$ sufficiently large we have$${4^ncdot (n!)^2over (2n)!}sim sqrt{pi n}$$and therefore the series diverges.
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
D'Alembert criterion is inconclusive beacause
$frac{a_{n+1}}{a_n}=frac{frac{4^{n+1};(n+1)!;(n+1)!}{left(2(n+1)right)!}}{frac{4^{n};n!;n!}{left(2nright)!}}=frac{4(n+1)(n+1)}{(2n+1)(2n+2)} rightarrow 1$ as $n rightarrow infty$, but as @Raymond Manzoni said, we could use Raabe's test. We are calculating $nleft(frac{a_n}{a_{n+1}}-1right)$. We have
begin{equation*}
nleft(frac{frac{4^n;n!;n!}{(2n)!}}{frac{4^{n+1};(n+1)!;(n+1)!}{(2n+2)!}}-1right)=nleft(frac{(2n+1)(2n+2)}{4(n+1)^2}-1right)=frac{-n^2-n}{2n^2+4n+4} \ =frac{-n(n+1)}{2(n+1)^2}leq 0,
end{equation*}
for all $n in mathbb{N}$ beacause the numerator is always negative and denominator positive, so it is especially less than $1$.
answered Jan 12 at 13:51
trulekstruleks
2114
$endgroup$
$begingroup$
Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
$endgroup$
– DiDo
Jan 12 at 13:57
$begingroup$
(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
$endgroup$
– Raymond Manzoni
Jan 12 at 14:12
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Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
$endgroup$
– truleks
Jan 12 at 14:16
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@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
$endgroup$
– Raymond Manzoni
Jan 12 at 14:19
1
$begingroup$
Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
$endgroup$
– DiDo
Jan 12 at 15:01
|
show 3 more comments
$begingroup$
D'Alembert criterion is inconclusive beacause
$frac{a_{n+1}}{a_n}=frac{frac{4^{n+1};(n+1)!;(n+1)!}{left(2(n+1)right)!}}{frac{4^{n};n!;n!}{left(2nright)!}}=frac{4(n+1)(n+1)}{(2n+1)(2n+2)} rightarrow 1$ as $n rightarrow infty$, but as @Raymond Manzoni said, we could use Raabe's test. We are calculating $nleft(frac{a_n}{a_{n+1}}-1right)$. We have
begin{equation*}
nleft(frac{frac{4^n;n!;n!}{(2n)!}}{frac{4^{n+1};(n+1)!;(n+1)!}{(2n+2)!}}-1right)=nleft(frac{(2n+1)(2n+2)}{4(n+1)^2}-1right)=frac{-n^2-n}{2n^2+4n+4} \ =frac{-n(n+1)}{2(n+1)^2}leq 0,
end{equation*}
for all $n in mathbb{N}$ beacause the numerator is always negative and denominator positive, so it is especially less than $1$.
answered Jan 12 at 13:51
trulekstruleks
2114
$endgroup$
$begingroup$
Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
$endgroup$
– DiDo
Jan 12 at 13:57
$begingroup$
(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
$endgroup$
– Raymond Manzoni
Jan 12 at 14:12
$begingroup$
Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
$endgroup$
– truleks
Jan 12 at 14:16
$begingroup$
@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
$endgroup$
– Raymond Manzoni
Jan 12 at 14:19
1
$begingroup$
Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
$endgroup$
– DiDo
Jan 12 at 15:01
|
show 3 more comments
$begingroup$
D'Alembert criterion is inconclusive beacause
$frac{a_{n+1}}{a_n}=frac{frac{4^{n+1};(n+1)!;(n+1)!}{left(2(n+1)right)!}}{frac{4^{n};n!;n!}{left(2nright)!}}=frac{4(n+1)(n+1)}{(2n+1)(2n+2)} rightarrow 1$ as $n rightarrow infty$, but as @Raymond Manzoni said, we could use Raabe's test. We are calculating $nleft(frac{a_n}{a_{n+1}}-1right)$. We have
begin{equation*}
nleft(frac{frac{4^n;n!;n!}{(2n)!}}{frac{4^{n+1};(n+1)!;(n+1)!}{(2n+2)!}}-1right)=nleft(frac{(2n+1)(2n+2)}{4(n+1)^2}-1right)=frac{-n^2-n}{2n^2+4n+4} \ =frac{-n(n+1)}{2(n+1)^2}leq 0,
end{equation*}
for all $n in mathbb{N}$ beacause the numerator is always negative and denominator positive, so it is especially less than $1$.
answered Jan 12 at 13:51
trulekstruleks
2114
$endgroup$
D'Alembert criterion is inconclusive beacause
$frac{a_{n+1}}{a_n}=frac{frac{4^{n+1};(n+1)!;(n+1)!}{left(2(n+1)right)!}}{frac{4^{n};n!;n!}{left(2nright)!}}=frac{4(n+1)(n+1)}{(2n+1)(2n+2)} rightarrow 1$ as $n rightarrow infty$, but as @Raymond Manzoni said, we could use Raabe's test. We are calculating $nleft(frac{a_n}{a_{n+1}}-1right)$. We have
begin{equation*}
nleft(frac{frac{4^n;n!;n!}{(2n)!}}{frac{4^{n+1};(n+1)!;(n+1)!}{(2n+2)!}}-1right)=nleft(frac{(2n+1)(2n+2)}{4(n+1)^2}-1right)=frac{-n^2-n}{2n^2+4n+4} \ =frac{-n(n+1)}{2(n+1)^2}leq 0,
end{equation*}
for all $n in mathbb{N}$ beacause the numerator is always negative and denominator positive, so it is especially less than $1$.
answered Jan 12 at 13:51
trulekstruleks
2114
edited Jan 12 at 15:32
answered Jan 12 at 13:51
trulekstruleks
2114
answered Jan 12 at 13:51
trulekstruleks
2114
answered Jan 12 at 13:51
trulekstruleks
2114
2114
$begingroup$
Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
$endgroup$
– DiDo
Jan 12 at 13:57
$begingroup$
(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
$endgroup$
– Raymond Manzoni
Jan 12 at 14:12
$begingroup$
Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
$endgroup$
– truleks
Jan 12 at 14:16
$begingroup$
@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
$endgroup$
– Raymond Manzoni
Jan 12 at 14:19
1
$begingroup$
Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
$endgroup$
– DiDo
Jan 12 at 15:01
|
show 3 more comments
$begingroup$
Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
$endgroup$
– DiDo
Jan 12 at 13:57
$begingroup$
(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
$endgroup$
– Raymond Manzoni
Jan 12 at 14:12
$begingroup$
Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
$endgroup$
– truleks
Jan 12 at 14:16
$begingroup$
@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
$endgroup$
– Raymond Manzoni
Jan 12 at 14:19
1
$begingroup$
Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
$endgroup$
– DiDo
Jan 12 at 15:01
$begingroup$
Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
$endgroup$
– DiDo
Jan 12 at 13:57
$begingroup$
Thank you dear, But by this way the value of the limit is equal to $1$, which means inconclusive and we can not use Ratio Test. Is there any other idea pleas.
$endgroup$
– DiDo
Jan 12 at 13:57
$begingroup$
(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
$endgroup$
– Raymond Manzoni
Jan 12 at 14:12
$begingroup$
(note that the $dfrac{a_{n+1}}{a_n}$ method itself is fine if we consider the extension by Raabe-Duhamel or the ratio test )
$endgroup$
– Raymond Manzoni
Jan 12 at 14:12
$begingroup$
Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
$endgroup$
– truleks
Jan 12 at 14:16
$begingroup$
Oh, sorry. I skipped a step calculating in head. I'll delete it in a minute if nothing comes up.
$endgroup$
– truleks
Jan 12 at 14:16
$begingroup$
@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
$endgroup$
– Raymond Manzoni
Jan 12 at 14:19
$begingroup$
@truleks: you don't need too, edit simply your answer renaming it as "ratio test" in the case limit$=1$ as explained in my ratio test link (we have indeed $left|frac{a_{n+1}}{a_n}right|ge 1$ for all large $n$). Fine continuation here,
$endgroup$
– Raymond Manzoni
Jan 12 at 14:19
1
1
$begingroup$
Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
$endgroup$
– DiDo
Jan 12 at 15:01
$begingroup$
Dear @RaymondManzoni the last calculation is wrong because $$ nleft(frac{left(2n+1right)left(2n+2right)}{4left(n+1right)^2}−1right) neq frac{-n^2-n}{2n^2 +4n +4}$$ in all cases we get $1$ again. Thank you for introducing another ways also and your reply.
$endgroup$
– DiDo
Jan 12 at 15:01
|
show 3 more comments
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Consider
$$sum _{n=0}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}=sum _{n=0}^{p }frac{4^{n:}n!:n!}{left(2nright)!}+sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}$$
Now, consider
$$a_n=frac{4^{n:}n!:n!}{left(2nright)!}implies log(a_n)=n log(4)+2log(n!)-log((2n)!)$$ and use Stirling approximation. This would give, for large $n$
$$log(a_n)=frac{1}{2} left(log (pi )+log left({n}right)right)+Oleft(frac{1}{n}right)implies a_nsimeq {sqrt{pi n}}$$ which shows that
$$sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}to infty$$
Edit
In fact, you could easily show, that for any $n geq 1$, $a_n >sqrt{pi n}$ which makes
$$sum _{n=0}^{p}frac{4^{n:}n!:n!}{left(2nright)!}=2+sum _{n=1}^{p }frac{4^{n:}n!:n!}{left(2nright)!} > 2+sum _{n=1}^{p } sqrt{pi n}=2+sqrt{pi } H_p^{left(-frac{1}{2}right)}$$ where appears generalized harmonic numbers. Using the asymptotics
$$ H_p^{left(-frac{1}{2}right)}=frac{2 }{3}p^{3/2}+frac{1}{2}p^{1/2}+zeta
left(-frac{1}{2}right)+Oleft(frac{1}{p^{1/2}}right)$$
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
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Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
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– Raymond Manzoni
Jan 12 at 15:34
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@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
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– Claude Leibovici
Jan 12 at 15:37
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Consider
$$sum _{n=0}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}=sum _{n=0}^{p }frac{4^{n:}n!:n!}{left(2nright)!}+sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}$$
Now, consider
$$a_n=frac{4^{n:}n!:n!}{left(2nright)!}implies log(a_n)=n log(4)+2log(n!)-log((2n)!)$$ and use Stirling approximation. This would give, for large $n$
$$log(a_n)=frac{1}{2} left(log (pi )+log left({n}right)right)+Oleft(frac{1}{n}right)implies a_nsimeq {sqrt{pi n}}$$ which shows that
$$sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}to infty$$
Edit
In fact, you could easily show, that for any $n geq 1$, $a_n >sqrt{pi n}$ which makes
$$sum _{n=0}^{p}frac{4^{n:}n!:n!}{left(2nright)!}=2+sum _{n=1}^{p }frac{4^{n:}n!:n!}{left(2nright)!} > 2+sum _{n=1}^{p } sqrt{pi n}=2+sqrt{pi } H_p^{left(-frac{1}{2}right)}$$ where appears generalized harmonic numbers. Using the asymptotics
$$ H_p^{left(-frac{1}{2}right)}=frac{2 }{3}p^{3/2}+frac{1}{2}p^{1/2}+zeta
left(-frac{1}{2}right)+Oleft(frac{1}{p^{1/2}}right)$$
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
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Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
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– Raymond Manzoni
Jan 12 at 15:34
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@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
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– Claude Leibovici
Jan 12 at 15:37
add a comment |
$begingroup$
Consider
$$sum _{n=0}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}=sum _{n=0}^{p }frac{4^{n:}n!:n!}{left(2nright)!}+sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}$$
Now, consider
$$a_n=frac{4^{n:}n!:n!}{left(2nright)!}implies log(a_n)=n log(4)+2log(n!)-log((2n)!)$$ and use Stirling approximation. This would give, for large $n$
$$log(a_n)=frac{1}{2} left(log (pi )+log left({n}right)right)+Oleft(frac{1}{n}right)implies a_nsimeq {sqrt{pi n}}$$ which shows that
$$sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}to infty$$
Edit
In fact, you could easily show, that for any $n geq 1$, $a_n >sqrt{pi n}$ which makes
$$sum _{n=0}^{p}frac{4^{n:}n!:n!}{left(2nright)!}=2+sum _{n=1}^{p }frac{4^{n:}n!:n!}{left(2nright)!} > 2+sum _{n=1}^{p } sqrt{pi n}=2+sqrt{pi } H_p^{left(-frac{1}{2}right)}$$ where appears generalized harmonic numbers. Using the asymptotics
$$ H_p^{left(-frac{1}{2}right)}=frac{2 }{3}p^{3/2}+frac{1}{2}p^{1/2}+zeta
left(-frac{1}{2}right)+Oleft(frac{1}{p^{1/2}}right)$$
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
$endgroup$
Consider
$$sum _{n=0}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}=sum _{n=0}^{p }frac{4^{n:}n!:n!}{left(2nright)!}+sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}$$
Now, consider
$$a_n=frac{4^{n:}n!:n!}{left(2nright)!}implies log(a_n)=n log(4)+2log(n!)-log((2n)!)$$ and use Stirling approximation. This would give, for large $n$
$$log(a_n)=frac{1}{2} left(log (pi )+log left({n}right)right)+Oleft(frac{1}{n}right)implies a_nsimeq {sqrt{pi n}}$$ which shows that
$$sum _{n=p+1}^{infty }frac{4^{n:}n!:n!}{left(2nright)!}to infty$$
Edit
In fact, you could easily show, that for any $n geq 1$, $a_n >sqrt{pi n}$ which makes
$$sum _{n=0}^{p}frac{4^{n:}n!:n!}{left(2nright)!}=2+sum _{n=1}^{p }frac{4^{n:}n!:n!}{left(2nright)!} > 2+sum _{n=1}^{p } sqrt{pi n}=2+sqrt{pi } H_p^{left(-frac{1}{2}right)}$$ where appears generalized harmonic numbers. Using the asymptotics
$$ H_p^{left(-frac{1}{2}right)}=frac{2 }{3}p^{3/2}+frac{1}{2}p^{1/2}+zeta
left(-frac{1}{2}right)+Oleft(frac{1}{p^{1/2}}right)$$
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
edited Jan 12 at 15:59
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
answered Jan 12 at 15:30
Claude LeiboviciClaude Leibovici
121k1157133
121k1157133
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Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
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– Raymond Manzoni
Jan 12 at 15:34
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@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
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– Claude Leibovici
Jan 12 at 15:37
add a comment |
$begingroup$
Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
$endgroup$
– Raymond Manzoni
Jan 12 at 15:34
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@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
$endgroup$
– Claude Leibovici
Jan 12 at 15:37
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Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
$endgroup$
– Raymond Manzoni
Jan 12 at 15:34
$begingroup$
Hi Claude! Nice to let you try it using Stirling (one of my old favorite). Just a little correction $a_napprox sqrt{pi n}$ (one more exponentiation he he). Cheers,
$endgroup$
– Raymond Manzoni
Jan 12 at 15:34
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@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
$endgroup$
– Claude Leibovici
Jan 12 at 15:37
$begingroup$
@RaymondManzoni. Hi Raymond ! Thanks for pointing the typo. Cheers.
$endgroup$
– Claude Leibovici
Jan 12 at 15:37
add a comment |
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Using Stirling's approximation for factorial and for large enough $n$ we have $${4^ncdot (n!)^2over (2n)!}{sim{4^n left[sqrt{2pi n}left({nover e}right)^nright]^2over {sqrt{4pi n}}left({2nover e}right)^{2n}}\={4^ncdot {2pi n}cdotleft({nover e}right)^{2n}over {sqrt{4pi n}}left({2nover e}right)^{2n}}\= {4^ncdot {2pi n}over {sqrt{4pi n}}cdot {4^n}}\=sqrt{pi n}}$$therefore for $n$ sufficiently large we have$${4^ncdot (n!)^2over (2n)!}sim sqrt{pi n}$$and therefore the series diverges.
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
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add a comment |
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Using Stirling's approximation for factorial and for large enough $n$ we have $${4^ncdot (n!)^2over (2n)!}{sim{4^n left[sqrt{2pi n}left({nover e}right)^nright]^2over {sqrt{4pi n}}left({2nover e}right)^{2n}}\={4^ncdot {2pi n}cdotleft({nover e}right)^{2n}over {sqrt{4pi n}}left({2nover e}right)^{2n}}\= {4^ncdot {2pi n}over {sqrt{4pi n}}cdot {4^n}}\=sqrt{pi n}}$$therefore for $n$ sufficiently large we have$${4^ncdot (n!)^2over (2n)!}sim sqrt{pi n}$$and therefore the series diverges.
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Using Stirling's approximation for factorial and for large enough $n$ we have $${4^ncdot (n!)^2over (2n)!}{sim{4^n left[sqrt{2pi n}left({nover e}right)^nright]^2over {sqrt{4pi n}}left({2nover e}right)^{2n}}\={4^ncdot {2pi n}cdotleft({nover e}right)^{2n}over {sqrt{4pi n}}left({2nover e}right)^{2n}}\= {4^ncdot {2pi n}over {sqrt{4pi n}}cdot {4^n}}\=sqrt{pi n}}$$therefore for $n$ sufficiently large we have$${4^ncdot (n!)^2over (2n)!}sim sqrt{pi n}$$and therefore the series diverges.
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Using Stirling's approximation for factorial and for large enough $n$ we have $${4^ncdot (n!)^2over (2n)!}{sim{4^n left[sqrt{2pi n}left({nover e}right)^nright]^2over {sqrt{4pi n}}left({2nover e}right)^{2n}}\={4^ncdot {2pi n}cdotleft({nover e}right)^{2n}over {sqrt{4pi n}}left({2nover e}right)^{2n}}\= {4^ncdot {2pi n}over {sqrt{4pi n}}cdot {4^n}}\=sqrt{pi n}}$$therefore for $n$ sufficiently large we have$${4^ncdot (n!)^2over (2n)!}sim sqrt{pi n}$$and therefore the series diverges.
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 12 at 20:21
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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Hint: use Stirling formula for the factorial (you should obtain the Wallis product)
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– Raymond Manzoni
Jan 12 at 13:44
1
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If $a_n$ is as in the question, I think $a_n to infty$, and therefore $sum a_n$ diverges.
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– GEdgar
Jan 12 at 14:28