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.
 A: $$f(x)=(1-x^2)^n=\sum_{k=0}^n \binom{n}{k}x^{2k}(-1)^k$$
$$F(n)=\int_0^1(1-x^2)^n dx=\sum_{k=0}^n \frac{(-1)^k}{2k+1} \binom nk\tag{1}$$
On the other side (see proof here):
$$F(n)=\int_{0}^{1}(1-x^2)^ndx={(2n)!!\over (2n+1)!!}$$
It's a simple exercise to show that:
$${(2n)!!\over (2n+1)!!}=\frac{4^n}{(2n+1){2n\choose n}}$$
...which completes the proof.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\sum_{k = 0}^{n}{\pars{-1}^{k} \over 2k + 1}{n \choose k} =
{4^{n} \over \pars{2n + 1}{2n\choose n}}:\ {\LARGE ?}}$.

\begin{align}
\sum_{k = 0}^{n}{\pars{-1}^{k} \over 2k + 1}{n \choose k} & =
\sum_{k = 0}^{n}\pars{-1}^{k}\pars{\int_{0}^{1}t^{2k}\dd t}{n \choose k} =
\int_{0}^{1}\bracks{\sum_{k = 0}^{n}{n \choose k}\pars{-t^{2}}^{k}}\dd t
\\[5mm] & =
\int_{0}^{1}\pars{1 - t^{2}}^{n}\dd t
\\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\,
{1 \over 2}\
\overbrace{\int_{0}^{1}t^{-1/2}\,\pars{1 - t}^{n}\,\dd t}
^{\ds{\mrm{B}\pars{1/2,n + 1}}}\qquad\pars{~\mrm{B}:\ Beta\ Function~}
\\[5mm] & =
{1 \over 2}\,{\Gamma\pars{1/2}\Gamma\pars{n + 1} \over \Gamma\pars{n + 3/2}}\qquad\pars{~\Gamma:\ Gamma\ Function~}
\\[5mm] & =
{1 \over 2}\,{\root{\pi}\Gamma\pars{n + 1} \over
\root{2\pi}2^{-3/2 - 2n}\Gamma\pars{2n + 2}/
\Gamma\pars{n + 1}}\label{1}\tag{1}
\end{align}
In the last expression, I used $\ds{\Gamma\pars{1/2} = \root{\pi}}$ and the Gamma Duplication Formula.

\eqref{1} becomes:

\begin{align}
\sum_{k = 0}^{n}{\pars{-1}^{k} \over 2k + 1}{n \choose k} & =
{4^{n}\pars{n!}^{2} \over \pars{2 n + 1}!} =
{4^{n} \over \pars{2n + 1}\bracks{\pars{2n}!/\pars{n!}^{2}}} =
\bbx{{4^{n} \over \pars{2n + 1}{2n \choose n}}}
\end{align}
A: 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.
A: I was actually thinking about exactly this sum a few days ago, and found a few ways to evaluate it. I'm on my phone right now so I can't tpe it up all pretty, but I'll give the general idea.
Method 1
(-1)^k / (2k+1) = 1/i * integral from 0 to i of x^2k dx
Multiplying by binomial coefficients gives you that the sum you're looking for is
integral from 0 to i of (1+x^2)^n dx
A simple integration by parts proves your formula via induction.
Method 2
1 / (k + 1/2) = integral from 0 to 1 of x^(k-1/2) dx
Multiplying by binomial coefficients and summing you get that the sum is equal to half of
integral from 0 to 1 of (1-x)^n * x^(-1/2) dx
But that's the beta function, and if you evaluate it using the formula involving the gamma function you get the correct answer.
Notice that method 2 easily generalizes to the case where we replace 2k+1 with xk+y where x and y are arbitrary real numbers, and indeed to the case where n is any real number.
