Evaluating $\sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$

$\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}=\sum_{n=i}^{\infty} \frac {1}{{2n \choose n-i}}$ 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).

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 $$ \sum_{n=1}^\infty\frac1{\binom{2n}{n-1}}=\frac13+\frac{5\pi}{9\sqrt3}\tag{11} $$

$$\sum_{n=0}^{+\infty}\binom{2n}{n}^{-1}=\sum_{n\geq 0}(2n+1)\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\sum_{n\geq 0}\int_{0}^{1}(2n+1)(x(1-x))^n\,dx$$ hence: $$\sum_{n=0}^{+\infty}\binom{2n}{n}^{-1}=\int_{0}^{1}\frac{1+x(1-x)}{(1-x(1-x))^2}\,dx$$ and the integral can be easily evaluated through the residue theorem. Other cases are similar.