The summation $\sum_{k=0}^n \frac{{2n \choose k}}{k+1}$

Question

Mathematica gives this summation
$$\sum_{k=0}^n \frac{{2n \choose k}}{k+1}$$
in terms of a regularized Gauss hypergeometric function $~_2F_1$. Can the hypergeometric function be eliminated to have a simple answer?

Answer 1

If you're looking for a (somewhat) closed form solution, observe that

\begin{aligned} \frac{\binom{2n}{k}}{k+1} &= \frac{(2n)!}{(k+1)! \cdot (2n - k)!} \\ &= \frac{1}{2n+1} \frac{(2n+1)!}{(k+1)! \cdot ((2n+1) - (k+1))!} \\ &= \frac{1}{2n+1}\binom{2n+1}{k+1}. \end{aligned}

So, by the Binomial theorem, we have

\begin{aligned} \sum_{k=0}^n \frac{\binom{2n}{k}}{k+1} &= \frac{1}{2n+1} \sum_{k=0}^n \binom{2n+1}{k+1} \\ &= \frac{1}{2n+1} \left(\sum_{k=0}^{n+1} \binom{2n+1}{k} - 1 \right) \\ &= \frac{1}{2n+1} \left(\frac{2^{2n+1}}{2} + \binom{2n+1}{n+1} - 1 \right) \\ &= \frac{1}{2n+1} \left(2^{2n} + \binom{2n+1}{n}- 1 \right). \end{aligned}