Finite series that gives Beta function at integers

Question

Playing around with Mathematica I have found that $$ \sum_{k=0}^n \binom{n}{k} \frac{(-1)^{n+k}}{2n+1-k} = \beta(n+1,n+1), $$ where $\beta(x,y)$ is the Beta function. Now, I'm not versed neither with series of this form, nor with the Beta function. Can someone help me see how this identity arises?

Answer 1

$$\begin{eqnarray*}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{n+k}}{2n+1-k} &=& \sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k}\int_{0}^{1} x^{2n-k}\,dx\\[0.2cm]&=&\int_{0}^{1}(-1)^n x^n\sum_{k=0}^{n}\binom{n}{k}(-1)^k x^{n-k}\,dx\\[0.2cm] &=& \int_{0}^{1}(-1)^n x^n (x-1)^n\,dx\\[0.2cm]&=&\int_{0}^{1}x^n (1-x)^n\,dx\\[0.2cm]&=& B(n+1,n+1)\\[0.2cm]&=&\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\\[0.2cm]&=&\frac{n!^2}{(2n+1)!}=\color{red}{\frac{1}{(2n+1)\binom{2n}{n}}}. \end{eqnarray*}$$