Sum of reciprocal binomial coefficients I am aware that 
$$\sum_{n=0}^\infty \binom{2n}{n}^{-1} = \frac{4}{3} + \frac{2\pi\sqrt{3}}{27}$$ 
though I do not know why it is true.  More generally, I'm interested in the value of the series 
$$S_k = \sum_{n=k}^\infty \binom{2n}{n-k}^{-1}$$ 
where $k$ is a fixed positive integer.  The series converges by the ratio test.  Does anybody know how to evaluate these sums, or have a reference where they are evaluated? 
 A: Maple writes your sum as a hypergeometric function:
$$ S_k = {\mbox{$_3$F$_2$}(1,1,1+2\,k;\,k+1,1/2+k;\,1/4)}$$
The values for $k=0$ to $3$ are
$$ \eqalign{k = 0: &{\frac{4}{3} + \frac {2\,\sqrt {3}\pi}{27}}
\cr
k = 1: &\frac13+{\frac {5\,\sqrt {3}\pi}{27}}
\cr
k = 2: &{\frac{23}{6}}-{\frac {13\,\sqrt {3}\pi}{27}}
\cr
k = 3: &\frac34+{\frac {2\,\sqrt {3}\pi}{27}}
\cr
}$$
Maple doesn't give a closed form for $k=4$ and up.  But I can get $k=4$ this way.
With $n = k + m$, write the summand as
$$ \frac{(m+2k)!\; m!}{(2m+2k)!} = \frac{m!^2}{(2m)!} \prod_{j=1}^{2k} \frac{m+j}{2m+j}  = 2^{-k}  \frac{m!^2}{(2m)!} \frac{\prod_{i=k+1}^{2k} (m+i)}{\prod_{i=0}^{k-1} (2m + 2i+1)}$$
Expand the quotient of products in partial fractions as a constant plus a sum of coefficients over $2m+j$.  Then sum individually.
That gives me
$$ S_4 = -\frac{211}{60} + \frac{23 \sqrt{3}\pi}{27}$$
but no farther since Maple won't give a closed form for 
$$ \sum_{m=0}^\infty \frac{m!^2}{(2m)!\; (2m+9)}$$
However, I think it should be possible to get closed forms for these: stay tuned.... 
EDIT: OK, it seems that
$$ F(z) = \sum_{m=0}^\infty \frac{m!^2}{(2m)!} z^{2m} = \frac{4}{4-z^2} + \frac{4z}{(4-z^2)^{3/2}} \arcsin(z/2) \ \text{for} |z|<2$$
so that
$$ \sum_{m=0}^\infty \frac{m!^2}{(2m)!(2m+j)} = \int_0^1 F(z) z^{j-1}\; dz $$
and these can be done in closed form.  So this gives me, for example,
$$S_5 = \frac{6169}{840} - \frac{31 \sqrt{3} \pi}{27} $$ 
and
$$ S_6 = \frac{1709}{2520} + \frac{2 \sqrt{3} \pi}{27} $$
