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).
 A: $$\sum_{n=0}^{+\infty}\binom{2n}{n}^{-1}=\sum_{n\geq 0}(2n+1)\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\sum_{n\geq 0}\int_{0}^{1}(2n+1)(x(1-x))^n\,dx$$
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.
A: We can do an example to see how this calculation actually works.
Suppose we seek to evaluate
$$\sum_{n\ge 4} {2n\choose n-4}^{-1}.$$
This is
$$\sum_{n\ge 4} \frac{(n-4)! \times (n+4)!}{(2n)!}
= \sum_{n\ge 4} 
\frac{\Gamma(n-3) \times \Gamma(n+5)}{\Gamma(2n+1)}
\\ = \sum_{n\ge 4} (2n+1)
 \frac{\Gamma(n-3) \times \Gamma(n+5)}{\Gamma(2n+2)}
= \sum_{n\ge 4} (2n+1) \mathrm{B}(n+5, n-3).$$
Recall the beta function integral
$$\mathrm{B}(x,y)
= \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}} dt.$$
This gives for the sum the representation
$$\int_0^\infty \sum_{n\ge 4} (2n+1)
\frac{t^{n+4}}{(1+t)^{2n+2}} dt
= \int_0^\infty \frac{t^4}{(1+t)^2}
\sum_{n\ge 4} (2n+1)
\frac{t^{n}}{(1+t)^{2n}} dt
\\ = \int_0^\infty \frac{t^4}{(1+t)^2}
\times \frac{(9t^2+11t+9)t^4}{(1+t)^6(t^2+t+1)^2} dt
\\ = \int_0^\infty
\frac{(9t^2+11t+9)t^8}{(1+t)^8(t^2+t+1)^2} dt.$$
This integral can be evaluated by considering
$$f(z) = 
\log z \times \frac{(9z^2+11z+9)z^8}{(1+z)^8(z^2+z+1)^2}$$
evaluated on a keyhole contour with the slot on the positive real axis
and  the branch  cut of  the logarithm  also on  that  axis, traversed
counterclockwise.

There are four segments: $\Gamma_1$ just above the cut, $\Gamma_2$ the
large circle of  radius $R$, $\Gamma_3$ the segment  below the cut and
$\Gamma_4$ the small circle around the origin of radius $\epsilon$.

Using $$|\log(Me^{i\theta})| = |\log M + i\theta|
= \sqrt{(\log M)^2+\theta^2}$$
we obtain that the contribution along $\Gamma_2$ is 
$$2\pi R \times \log R / R^2 \to 0$$
as $R\to \infty,$ so it vanishes.

The contribution along $\Gamma_4$ is 
$$2\pi\epsilon \times |\log \epsilon| \times \epsilon^8 \to 0$$
as $\epsilon\to 0,$ so it vanishes as well.

We get  two contributions from  the logarithm below the  positive real
axis  along  $\Gamma_3$,  one  of  which cancels  the  integral  along
$\Gamma_1$ and the other one of which is
$$-2\pi i 
\int_0^\infty
\frac{(9t^2+11t+9)t^8}{(1+t)^8(t^2+t+1)^2} dt$$ 
i.e. the integral we are trying to calculate.

Let $\rho_0 = -1$ and 
$$\rho_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{3}i}{2}.$$
We thus have that by the Cauchy Residue Theorem applied to the keyhole
contour
$$\int_0^\infty
\frac{(9t^2+11t+9)t^8}{(1+t)^8(t^2+t+1)^2} dt
= - (\mathrm{Res}_{z=\rho_0} f(z)
+ \mathrm{Res}_{z=\rho_1} f(z) + \mathrm{Res}_{z=\rho_2} f(z)). $$
Now to  calculate these residues  we use a  CAS but it will  need some
assistance namely from the expansion about $\rho$
$$\log z= \log(\rho + z -\rho)
= \log\rho + \log (1 + (z-\rho)/\rho)
= \log\rho +
\sum_{q\ge 1} \frac{(-1)^{q+1}}{q} \frac{(z-\rho)^q}{\rho^q}.$$

We use this expansion to compute the residues, making sure that the 
constant term $\log\rho$ agrees with the chosen branch.
This gives for the first residue that
$$\mathrm{Res}_{z=\rho_0} f(z)
= \frac{57}{20} - 3\pi i$$
and for the second
$$\mathrm{Res}_{z=\rho_1} f(z)
= \frac{1}{3} + \frac{23\pi \sqrt{3}}{27}
+ \pi i -\frac{1}{3}\sqrt{3}i$$
and for the third
$$\mathrm{Res}_{z=\rho_2} f(z)
= \frac{1}{3} - \frac{46\pi \sqrt{3}}{27}
+ 2\pi i +\frac{1}{3}\sqrt{3}i.$$
Adding these and negating the result we finally have
$$\frac{23\pi\sqrt{3}}{27} -  \frac{211}{60}.$$
A: Using Maple, I also get
$$ \sum _{n=2}^{\infty }  {2\,n\choose n-2}  ^{-1}={\frac {
23}{6}}-{\frac {13\,\pi \,\sqrt {3}}{27}}
$$
$$\sum _{n=3}^{\infty }  {2\,n\choose n-3} ^{-1}=\dfrac34+{
\frac {2\,\pi \,\sqrt {3}}{27}}
$$
For $i=4$ Maple can't seem to simplify
$$\sum _{n=4}^{\infty }  {2\,n\choose n-4}  ^{-1}=
{\mbox{$_3$F$_2$}(1,1,9;\,9/2,5;\,1/4)}
$$
but numerically it appears that the answer is $$- \dfrac{211}{60}+\dfrac{23 \pi \sqrt{3}}{27}$$
for $i=5$ it is 
$$ \dfrac{6169}{840}-\dfrac{31 \pi \sqrt{3}}{27}$$
and for $i=6$ it is
$$ {\frac {1709}{2520}}+{\frac {2\,\pi \,\sqrt {3}}{27}}$$
etc.  The coefficient of $\sqrt{3} \pi/27$ seems to have the generating function
$${\frac {6\,t-6}{ \left( {t}^{2}+t+1 \right) ^{2}}}+{\frac {-5\,t+8}{{t
}^{2}+t+1}}
$$
EDIT:
Ah!  Using Jack's method, the generating function for this sequence turns out to be
$$ \eqalign{
\sum_{i=0}^\infty t^i \sum_{n=i}^\infty {2n \choose n-i}^{-1} &= 
\sum_{i=0}^\infty t^i \sum_{n=i}^\infty \int_0^1 (n+i) t^{n-i} (1-t)^{n+i-1}\; dt\cr
=&\dfrac{\ln((1+\sqrt{t})/(1-\sqrt{t})) (t+1) t^{3/2}}{(t^2+t+1)^2}
- \dfrac{ \ln(1-t) (t^2 + 3 t + 1) t}{2(t^2+t+1)^2}\cr
& - \dfrac{(5 t^3 - 3 t^2 - 9 t - 2) \pi \sqrt{3}}{27 (t^2 + t + 1)^2}
 + \dfrac{1-t}{3 (t^2 + t + 1)} - \dfrac{1}{t-1} - 1 }$$
A: The basic result is $~\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x.~$ By repeating the process of differentiation 
and multiplication twice, and letting $x=\dfrac12,~$ you will obtain a closed form for your series. 
Then, by using similar means, try deducing a general formula for each modified version of 
this initial result, for $i>0$, namely, $i=1,2,3$, etc. If WA does not return nice results, use 
FullSimplify[...].
