TITLE: Proof that $\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}{n\choose k}=\frac{4^n}{(2n+1){2n\choose n}}$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.