Prove $$\sum_{n=1}^\infty \frac{1}{n^k\binom{2n}{n}}=\frac12 {_{k+1}F_k}\left(\underbrace{1,\dots,1}_{k+1};\frac32,\underbrace{2,\dots,2}_{k-1};\frac14\right)$$
where ${_m F_n}$ is the generalized hypergeometric function (which is referenced in equation 44) using the partial fraction decomposition of the reciprocal binomial: $$\frac{1}{\binom{z}{n}} = \sum_{i=0}^{n-1}(-1)^{n-1-i}\binom{n}{i}\frac{n-i}{z-i}$$
replacing $z$ with $2n$. Or
$$\frac{1}{\binom{z+n}{n}} = \sum_{i=1}^{n}(-1)^{i-1}\binom{n}{i}\frac{i}{z+i}$$
replacing $z$ with $n$. The above identities are given here.
My motivation is, by learning how to do this, to see if the first sum can be generalized to non-central binomial coefficients using the partial fraction decomposition formula.