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First off, I would like to apologize again for the integral I posted several days ago involving $\zeta(5)$. I was careless and did not examine the decimals out far enough.

With that said, I would now like to post a series I think is interesting. I am trying to derive a general form for

$$ \sum_{n=1}^{\infty}\frac{nx^{n}}{\binom{2n}{n}}.$$

I thought about starting with $\displaystyle \sum_{n=1}^{\infty}\frac{2^{2n}x^{2n}}{\binom{2n}{n}}=\frac{x^{2}}{\sqrt{1-x^{2}}}+\frac{x\sin^{-1}(x)}{(1-x^{2})^{\frac{3}{2}}}$.

I tried differentiating, integrating and so forth, but it turns into a mess and I do not know how to eliminate the $2^{2n}$ nor get the $x^{2n}$ down to $x^{n}$. Is it possible to somehow integrate in terms of, say, $t$ from $0$ to $x$?

Any thoughts on how to go about this?. This would then lead to $\displaystyle \sum_{n=1}^{\infty}\frac{n2^{n}}{\binom{2n}{n}}=\pi +3$ and many other forms just by using a general formula.

I ran across this in "Irresistible Integrals" by Boros and Moll. It is one of their 'Exercises'.

Thanks very much.

Davide Giraudo
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Cody
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    You might want to look at Lehmer's Interesting Series by F.J.Dyson, N.E. Frankel, M.L. Glasser where they consider $S_k(z)=\sum_{m=1}^{\infty}\frac{m^kz^m}{2m \choose m}$ – Henry Nov 26 '11 at 13:53
  • It's a messy combination of square roots and inverse trigonometric functions. I can easily derive the hypergeometric form, but a quick way of finding the elementary expression would certainly be interesting. – J. M. ain't a mathematician Nov 26 '11 at 14:02
  • @Cody, I thought the question about $\zeta(5)$ was great, I'm glad you asked it because reading the answers was fun. –  Nov 26 '11 at 14:25
  • Thanks for the link. QED, when I first looked at that integral, I thought Wow...that's cool. I give the poster to came up with that kudoes for deriving that much himself with that much accuracy. It may not have been 'on the money', but it was very clever and accurate nonetheless. – Cody Nov 26 '11 at 16:00

2 Answers2

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Start with $$f(x) = \sum_{n=1}^{\infty}\frac{2^{2n}x^{2n}}{\binom{2n}{n}}$$ where you know that $ \displaystyle f(x) = \frac{x^{2}}{\sqrt{1-x^{2}}}+\frac{x\sin^{-1}(x)}{(1-x^{2})^{3/2}}.$ Then $$ f'(x) = \sum_{n=1}^{\infty} \frac{ 2^{2n} \cdot 2n \cdot x^{2n-1} }{\binom{2n}{n}}.$$

Multiplying both sides by $x/2$ gives $$ \frac{x f'(x) }{2} = \sum_{n=1}^{\infty} \frac{ n 2^{2n} x^{2n} }{ \binom{2n}{n}} .$$

Then let $ \displaystyle x=\frac{ \sqrt{z} }{2} $ so

$$ \frac{\sqrt{z} }{2\sqrt{2}} f'\left( \frac{\sqrt{z} }{2} \right) = \sum_{n=1}^{\infty} \frac{ n z^n}{\binom{2n}{n} } .$$

Carrying out the final computation gives the sum to be $$\frac{6z}{(z -4)^2} + \frac{ 4\sqrt{z} (z+2) \csc^{-1}(2z^{-1/2} ) }{\sqrt{4-z} (z-4)^2 } .$$

Ragib Zaman
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  • Thanks Ragib. I made it as far as the x/2. I did not think of making a sub. Nice. Thanks. – Cody Nov 26 '11 at 16:05
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{n = 1}^{\infty}{nx^{n} \over {2n \choose n}}:\ {\large ?}}$

\begin{align}&\color{#66f}{\large\sum_{n = 1}^{\infty}{nx^{n} \over {2n \choose n}}} =\sum_{n = 1}^{\infty}nx^{n}\, {\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} =\sum_{n = 1}^{\infty}n^{2}x^{n}\, {\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} \\[3mm]&=\sum_{n = 1}^{\infty}n^{2}x^{n} \int_{0}^{1}t^{n - 1}\pars{1 - t}^{n}\,\dd t =\int_{0}^{1}\sum_{n = 1}^{\infty}n^{2}\bracks{xt\pars{1 - t}}^{n}\,{\dd t \over t} \\[3mm]&=\int_{0}^{1} \frac{(t-1) x \left(t^2 x-t x-1\right)}{\left(t^2 x-t x+1\right)^3}\,\dd t \\[3mm]&=\color{#66f}{\large\frac{x \left[6 \root{(4 - x) x}+4 (x+2) \arctan\left(\root{x}/\root{4-x}\right)\right]}{(x-4)^2 \root{(4 - x) x}}} \end{align}

Felix Marin
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