Evaluating $\sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$
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.
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 $$\frac{23\pi\sqrt{3}}{27} - \frac{211}{60}.$$