Proof that $\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$ I saw in this paper the following identity: $$\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$$ I have a pervious post on an integral quite closely related to this identity, but I still do not know how to derive/prove the identity. I'm really not that good at combinatorics or evaluating series, so I don't know how to start, which is why I don't have any attempts to show you. Please explain your steps thoroughly. Edit: I know there other posts on this series, but I did not get from them the proof I was satisfied by.

The Binomial Theorem says $$ \sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}=\left(1-x^2\right)^n $$ Integrating both sides over $[0,1]$ gives $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k} &=\int_0^1\left(1-x^2\right)^n\,\mathrm{d}x\\\ &=\frac12\int_0^1\left(1-x\right)^nx^{-1/2}\,\mathrm{d}x\\\ &=\frac12\frac{\Gamma(n+1)\,\Gamma(1/2)}{\Gamma(n+3/2)}\\\ &=\frac12\frac{n!}{\frac12\frac32\cdots\left(n+\frac12\right)}\\\ &=\frac{2^nn!}{1\cdot3\cdots(2n+1)}\\\ &=\frac{2^nn!2^nn!}{(2n+1)!}\\\ &=\frac{4^n}{(2n+1)\binom{2n}{n}} \end{align} $$ Using the Beta Function.