Sum of reciprocal binomial coefficients

Generally similar sums can be evaluated using the Beta function: $$ B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}= \frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}. $$

Applying this in your case ($x=n-k,y=n+k$) one has: $$ \binom{2n}{n-k}^{-1}=(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt $$ or $$\begin{align} \sum_{n=k}^\infty\binom{2n}{n-k}^{-1} &=\sum_{n=k}^\infty(2n+1)\int_0^1 t^{n-k}(1-t)^{n+k}dt\\ &=\int_0^1 dt \sum_{n=k}^\infty(2n+1)t^{n-k}(1-t)^{n+k}\\ &=\int_0^1 \frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}dt\tag1 \end{align}$$

It can be shown that the integral $(1)$ is a sum of a rational number and a multiple of $\dfrac\pi{9\sqrt3}$.

Indeed: $$ \frac{(1-t)^{2k}[2+(2k-1)(1-t+t^2)]}{(1-t+t^2)^2}=Q_k(t)+\frac{A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3}{(1-t+t^2)^2},\tag2 $$ where both coefficients of the polynomial $Q(t)$ and $A^0,A^1,A^2,A^3$ are integer numbers. The integral of $Q(t)$ is obviously a rational number and $I_r= \int_0^1\frac{t^r\,dt}{(1-t+t^2)^2}$ can be evaluated as: $$ I_0=\frac23+\frac49\dfrac\pi{\sqrt3};\quad I_1=\frac13+\frac29\dfrac\pi{\sqrt3};\quad I_2=-\frac13+\frac49\dfrac\pi{\sqrt3};\quad I_3=-\frac23+\frac59\dfrac\pi{\sqrt3}.\quad $$

Thus the irrational term can be written as: $$ C_k\frac\pi{9\sqrt3}\quad\text{with}\quad C_k=4A_k^0+2A_k^1+4A_k^2+5A_k^3. $$

Moreover the term can be evaluated explicitly using the following table: $$ \begin{array}{c|c|c|c|c|c} k\mod 3& A_k^0&A_k^1&A_k^2&A_k^3&C_k\\ \hline 0&+1-2x&+1+6x&-1-6x&+0+4x&2\\ 1&+2+4x&-5-6x&+3+6x&-1-2x&\hphantom{-1}5+18x\\ 2&-3-2x&2&0&-1-2x&-13-18x\\ \hline \end{array},\tag3 $$ with $x=\left\lfloor\dfrac k3\right\rfloor$, so that $C_k=2,5,-13,2,23,-31,2,41,-49,2,\dots$ for $k=0,1,2,3,4,5,6,7,8,9,\dots$.

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The expression $(3)$ can be proved in the following way:

Let $$ P_k(t)=(1-t)^{2k}[2+(2k-1)(1-t+t^2)];\quad R_k(t)=A_k^0+A_k^1t+A_k^2t^2+A_k^3t^3.\\ $$

Then we have from (2): $$R_k(t_\pm)=P_k(t_\pm);\quad R'_k(t_\pm)=P'_k(t_\pm),\tag4$$ where $$ t_\pm=e^{\pm\frac{i\pi}3} $$ are the roots of the polynomial $t^2-t+1$.

Explicitly (4) amounts to the system of four linear equations: $$\begin{align} A_k^0+A_k^1t_\pm+A_k^2t_\pm^2+A_k^3t_\pm^3&=2t_\pm^{-2k}\\ A_k^1+2A_k^2t_\pm+3A_k^3t_\pm^2&=(1-2k-2t_\pm)t_\pm^{-2k}\\ \end{align}, $$ which solutions are given by (3).

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 $$ S_6 = \frac{1709}{2520} + \frac{2 \sqrt{3} \pi}{27} $$