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 July 5th, 2011, 03:27 PM #1 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Series with inverse binomial factor! [color=#000000]Compute the series $\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{\binom{2n}{n}}$.[/color]
 July 7th, 2011, 03:30 AM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Series with inverse binomial factor! $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\binom{2n}{n}}$ $-\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\binom{2n}{n}}$ $-\sum_{n=1}^{\infty}\frac{(-1)^{n}(n!)^{2}}{(2n)!}$ Use of the Gamma function: $-\sum_{n=1}^{\infty}\frac{(-1)^{n}n\Gamma(n+1)\Gamma(n)}{\Gamma(2n+1)}$ Now, use the Beta/Gamma relation: $-\sum_{n=1}^{\infty}\int_{0}^{1}(-1)^{n}nx^{n}(1-x)^{n-1}dx$ $-\int_{0}^{1}\left[\frac{1}{1-x}\sum_{n=1}^{\infty}n(x(1-x))^{n}(-1)^{n}\right]dx$ $\int_{0}^{1}\left[\frac{1}{x-1}\sum_{n=1}^{\infty}n(x^{2}-x)^{n}\right]dx$ Using the sum of a geometric series, differentiating, then integrating: $\int_{0}^{1}\frac{x}{(1+x-x^{2})^{2}}dx$ $\fbox{=\frac{4\sqrt{5}ln\left(\frac{\sqrt{5}+1}{2} \right)}{25}+\frac{1}{5} \;\ \approx \;\ .3721635...}$ What is interesting to notice is that the Golden Ratio appears in the evaluation.
 July 7th, 2011, 04:20 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Series with inverse binomial factor! [color=#000000] nice!! Now try $\sum_{n=1}^{+\infty}\frac{(-1)^{n}}{n\binom{2n}{n}}$.[/color]
 July 7th, 2011, 04:45 AM #4 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Series with inverse binomial factor! In that event, the same previous idea can be used except the integral becomes: $\int_{0}^{1}\frac{1-x}{-x^{2}+x+1}dx=\frac{ln\left(\frac{\sqrt{5}+3}{2}\ri ght)}{\sqrt{5}} \;\ \approx \;\ .4304....$ How about $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}$?. Or, maybe even change the power of the n to a 2 or whatever.
 July 7th, 2011, 04:50 AM #5 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Series with inverse binomial factor! [color=#000000]I have the solution, for $n^2$ too, but for the 3rd power or higher no, you give some good ideas...[/color]
 July 7th, 2011, 05:35 AM #6 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Series with inverse binomial factor! deleted
July 8th, 2011, 03:59 PM   #7
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Re: Series with inverse binomial factor!

Quote:
 Originally Posted by galactus How about $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}$?.
$\frac{2}{5}\zeta(3)?$

July 9th, 2011, 03:37 AM   #8
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Re: Series with inverse binomial factor!

Quote:
Originally Posted by Stupid_man
Quote:
 Originally Posted by galactus How about $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}$?.
$\frac{2}{5}\zeta(3)?$

It involves the use of the polylogarithm.

By using the Beta function:

$\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}=\frac{(-1)^{n-1}(n!)^{2}}{n^{3}\binom{2n}{n}}=\frac{(-1)^{n-1}}{n^{3}}(2n+1)\beta(n+1,n+1)$

So, $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}=-\int_{0}^{1}\sum_{n=1}^{\infty}\frac{(2n+1)(x^{2}-x)^{n}}{n^{3}}dx$

By using the polylog: $Li_{3}(x)=\sum_{n=1}^{\infty}\frac{x^{n}}{n^{3}}$, we get:

$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}=-\int_{0}^{1}\left[2Li_{2}(x^{2}-x)+Li_{3}(x^{2}-x)\right]dx=\frac{2}{5}\zeta(3)$

Integration by parts can be used, along with a little knowledge of the polylog, to arrive at the sum. It's rather tedious.

If you have another way, I am always open to seeing other methods.

The best book on the topic is "Polylogarithms and associated functions" by Leonard Lewin.

 September 6th, 2011, 04:31 PM #9 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Series with inverse binomial factor! [color=#000000]Galactus I remember we have forgotten something after a long time.......took me months to finally solve it! Computation of $\color{blue}\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n^2\binom{2n}{n}}$. $\color{red}\rule{500pt}{1pt}$ $\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}x^n}{n\binom{2n}{n}}=\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}(n!)^{2}x^n}{n(2n!)^{2}}=\sum_{n=1}^{+\infty}\fr ac{(-1)^{n-1}x^{n}n\Gamma(n+1)\Gamma(n)}{n\Gamma(2n+1)}=\sum_ {n=1}^{+\infty}(-1)^{n-1}x^{n}B(n+1,n)=$ $\sum_{n=1}^{+\infty}(-1)^{n-1}x^{n}\int_{0}^{1}y^{n}(1-y)^{n-1}\;dy=\int_{0}^{1}xy\left[\sum_{n=0}^{+\infty}\left(xy(y-1)\right)^{n}\right]\;dy=-\int_{0}^{1}\frac{y}{y^{2}-y-\frac{1}{x}}\;dy=-\frac{2\sqrt{x}}{\sqrt{4+x}}\ln\left(\frac{\sqrt{4 +x}-\sqrt{x}}{2}\right)$ and so $\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}x^{n}}{n\binom{2n}{n}}=-\frac{2\sqrt{x}}{\sqrt{4+x}}\ln\left(\frac{\sqrt{4 +x}-\sqrt{x}}{2}\right)\Rightarrow\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}x^{n-1}}{n\binom{2n}{n}}=-\frac{2}{\sqrt{x}\sqrt{4+x}}\ln\left(\frac{\sqrt{4 +x}-\sqrt{x}}{2}\right)\Rightarrow$ $\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n\binom{2n}{n}}\int_{0}^{x}w^{n-1}\;dw=-2\int_{0}^{x}\frac{\ln\left(\frac{\sqrt{4+w}-\sqrt{w}}{2}\right)}{\sqrt{w}\sqrt{4+w}}\;dw\Right arrow \fbox{\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}x^n}{n^{2}\binom{2n}{n}}=-2\int_{0}^{x}\frac{\ln\left(\frac{\sqrt{4+w}-\sqrt{w}}{2}\right)}{\sqrt{w}\sqrt{4+w}}\;dw$ hence for x=1 we get, $\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n^2\binom{2n}{n}}=-2\int_{0}^{1}\frac{\ln\left(\frac{\sqrt{4+w}-\sqrt{w}}{2}\right)}{\sqrt{w}\sqrt{4+w}}\;dw\;\;\; \overset{x=\ln\left(\frac{\sqrt{4+w}-\sqrt{w}}{2}\right)}{=}\;\;\;2\ln^{2}\left(\frac{1 +\sqrt{5}}{2}\right)$ .[/color]
 September 17th, 2011, 12:28 PM #10 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: Series with inverse binomial factor! Very nice, Z. That's a clever approach. Good ol' Beta and Gamma sure come in handy, huh?. Sorry I have not been around for a while. Been very busy and have not been on the sites much lately. If you're interested, and I think you are, here is a general form for a series: $\sum_{k=1}^{\infty}\frac{1}{k^{2}\binom{mk}{m}}=-(m-1)\int_{0}^{1}\frac{ln(1-x^{m-1}(1-x))}{x}dx$ If m=2, then we have $-\int_{0}^{1}\frac{ln(1-x(1-x))}{x}dx$ If m=3, then we have $-2\int_{0}^{1}\frac{ln(1-x^{2}(1-x))}{x}dx$ and so on. The integration is not too elementary, though.

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