Derive a closed form for a sum with inverse binomial coefficients

Question

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.

Answer 1

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 $$\frac{6z}{(z -4)^2} + \frac{ 4\sqrt{z} (z+2) \csc^{-1}(2z^{-1/2} ) }{\sqrt{4-z} (z-4)^2 } .$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $$ \sum_{n = 1}^{\infty}{nx^{n} \over \ds{2n \choose n}}:\ {\LARGE ?} $$

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\begin{align} & \color{#66f}{\sum_{n = 1}^{\infty}{nx^{n} \over \ds{2n \choose n}}} = \sum_{n = 1}^{\infty}nx^{n}\, {\Gamma\pars{n + 1}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} \\[5mm] = & \ \sum_{n = 1}^{\infty}n^{2}x^{n}\, {\Gamma\pars{n}\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}} = \sum_{n = 1}^{\infty}n^{2}x^{n} \int_{0}^{1}t^{n - 1}\,\pars{1 - t}^{n}\,\dd t \\[5mm] = & \ \int_{0}^{1}\sum_{n = 1}^{\infty} n^{2}\bracks{xt\pars{1 - t}}^{n} \,{\dd t \over t} = \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 \\[5mm] = & \ \bbx{\!\!\!\!\color{#66f}{\frac{x\bracks{6 \root{\pars{4 - x}x} + 4\pars{x + 2} \arctan\pars{\root{x}/\root{4-x}}}} {\pars{x - 4}^{2}\root{\pars{4 - x}x}}}\!\!\!} \\ & \end{align}