TITLE: Evaluating $\sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$
QUESTION [13 upvotes]: $\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}=\sum_{n=i}^{\infty} \frac {1}{{2n \choose n-i}}$ is a very interesting one. Here is what I have from WolframAlpha.
$$\displaystyle \sum_{n=0}^{\infty} {2n \choose n}^{-1}=\frac{2}{27}(18+\sqrt{3}\pi)$$
$$\displaystyle \sum_{n=1}^{\infty} {2n \choose n}^{-1}=\frac{1}{27}(9+2\sqrt{3}\pi)$$
$$\displaystyle \sum_{n=1}^{\infty} {2n \choose n-1}^{-1}=\frac{1}{27}(9+5\sqrt{3}\pi)$$
For $i \geq 2$, WA just comes up with closed forms involving generalised hypergeometric functions. Here is one example. I would conjecture that this is still in fact expressible in terms of the sort of form we see above but I have no clue about hypergeometric functions so I was hoping somebody could enlighten me. Also, it looks like some sort of computational artefact of not being able to start the sum at $n=1$. It seems to give a natural, consistent form for the first two cases ($i=0,1$), and when asking WA to sum it in the case $i=1$, but starting from $n=2$, it uses a hypergeometric function in the answer, whereas we know the answer should in fact be $\frac{1}{27}(-18+5\sqrt{3}\pi)$.
Also, could anybody come up with a (more elementary i.e. anything not involving special functions) proof of the formula
$\displaystyle \sum_{n=1}^{\infty} {2n \choose n}^{-1}=\frac{1}{27}(9+2\sqrt{3}\pi)$?
I have struggled and failed, and this is the part I would most like answered if possible.
Thanks, and good luck (if needed).

REPLY [7 votes]: Generating Function for the Reciprocals of the Central Binomial Coefficients
Using the fact that
$$
\frac{4^n}{\binom{2n}{n}}
=\frac{n}{n-\frac12}\frac{4^{n-1}}{\binom{2n-2}{n-1}}\tag{1}
$$
we can compute the generating function for the reciprocals of the central binomial coefficients:
$$
\begin{align}
\frac{4^n}{\binom{2n}{n}}
&=\prod_{k=1}^n\frac{k}{k-\frac12}\\
&=\frac{\Gamma(n+1)}{\Gamma(1)}\frac{\Gamma(\frac12)}{\Gamma(n+\frac12)}\\
&=n\,\mathrm{B}(n,\tfrac12)\\
&=n\int_0^1(1-t)^{n-1}t^{-1/2}\,\mathrm{d}t\\
&=2n\int_0^1(1-t^2)^{n-1}\,\mathrm{d}t\\
\frac{x^n}{\binom{2n}{n}}
&=\frac{nx}2\int_0^1\left(\frac{x(1-t^2)}4\right)^{n-1}\,\mathrm{d}t\\
\sum_{n=1}^\infty\frac{x^n}{\binom{2n}{n}}
&=\frac{x}2\int_0^1\frac1{\left(1-\frac{x(1-t^2)}4\right)^2}\,\mathrm{d}t\\
&=\frac{8x}{(4-x)^2}\int_0^1\frac1{\left(1+\frac{x}{4-x}t^2\right)^2}\,\mathrm{d}t\\
&=\frac{8x^{1/2}}{(4-x)^{3/2}}\int_0^{\sqrt{\frac{x}{4-x}}}\frac1{\left(1+t^2\right)^2}\,\mathrm{d}t\\
&=\frac{4x^{1/2}}{(4-x)^{3/2}}\left[\frac{t}{1+t^2}+\tan^{-1}(t)\right]_0^{\sqrt{\frac{x}{4-x}}}\\
&=\frac{x}{4-x}+\frac4{4-x}\sqrt{\frac{x}{4-x}}\tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right)\\
\sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}}
&=\frac4{4-x}\left[1+\sqrt{\frac{x}{4-x}}\tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right)\right]\\
&=\frac4{4-x}\left[1+\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\right]\tag{2}
\end{align}
$$
Plug in $x=1$ and we get
$$
\begin{align}
\sum_{n=0}^\infty\frac1{\binom{2n}{n}}
&=\frac43\left[1+\sqrt{\frac13}\sin^{-1}\left(\frac12\right)\right]\\
&=\frac43+\frac{2\pi}{9\sqrt3}\tag{3}
\end{align}
$$

Extending the Previous Result
Suppose we have
$$
F_k(x)=\sum_{n=k}^\infty\frac{x^{n-k}}{\binom{2n}{n-k}}\tag{4}
$$
$F_0(x)$ is given in $(2)$ above.
We can use the identity
$$
\binom{2n}{n-k-1}=\frac{n-k}{n+k+1}\binom{2n}{n-k}\tag{5}
$$
to get
$$
\begin{align}
\frac1{\binom{2n}{n-k-1}}-\frac1{\binom{2n}{n-k}}
&=\frac1{\frac{n-k}{n+k+1}\binom{2n}{n-k}}-\frac1{\binom{2n}{n-k}}\\
&=\frac1{n-k}\left[\frac{n+k+1}{\binom{2n}{n-k}}-\frac{n-k}{\binom{2n}{n-k}}\right]\\
&=\frac{2k+1}{n-k}\frac1{\binom{2n}{n-k}}\tag{6}
\end{align}
$$
Equation $(6)$ shows that
$$
\frac{\mathrm{d}}{\mathrm{d}x}(xF_{k+1}(x)-F_k(x))=\frac{2k+1}{x}(F_k(x)-1)\tag{7}
$$
Formula $(7)$ gives us a way to compute $F_{k+1}$ from $F_k$, via integration.

Example
If we let $x=4\sin^2(\theta)$, then
$$
\begin{align}
&\int\frac{F_0(x)-1}{x}\mathrm{d}x\\
&=\int\left[\frac1{4-x}+\frac4{4-x}\sqrt{\frac1{x(4-x)}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\right]\mathrm{d}x\\
&=\int\left[\frac1{4\cos^2(\theta)}+\frac1{\cos^2(\theta)}\sqrt{\frac1{16\sin^2(\theta)\cos^2(\theta)}}\,\,\theta\right]8\sin(\theta)\cos(\theta)\,\mathrm{d}\theta\\[3pt]
&=2\int\left[\tan(\theta)+\theta\sec^2(\theta)\right]\mathrm{d}\theta\\[12pt]
&=2\theta\tan(\theta)-1\\[9pt]
&=2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)-1\tag{8}
\end{align}
$$
where the constant of integration was chosen because $xF_1(x)-F_0(x)=-1$ at $x=0$.
Thus, we get
$$
xF_1(x)-F_0(x)=2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)-1\tag{9}
$$
Therefore,
$$
F_1(x)=\frac1{4-x}+\frac{12-2x}{4-x}\frac1{\sqrt{x(4-x)}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\tag{10}
$$
Plug in $x=1$ and we get
$$
\sum_{n=1}^\infty\frac1{\binom{2n}{n-1}}=\frac13+\frac{5\pi}{9\sqrt3}\tag{11}
$$