Closed form of sum with binomial

Question

I want to find closed form of the following expression :

$$\sum\limits_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{2k+1}$$

I have no idea how to do it.

Answer 1

Hint: Note that \begin{align*} \sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{2k+1} &= \int_0^1 \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{2k} \, dx \end{align*} and $$ \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{2k} = \sum_{k=0}^{n} \binom{n}{k} (-x^2)^k \, . $$ Do you see how to finish this using the binomial theorem? This will allow you to express the sum in terms of the beta function.

Answer 2

Let $$ f(x)=\sum\limits_{k=0}^{n} \binom{n}{k}\frac{(-1)^k}{2k+1}x^{2k+1}.$$ Then calculate $f'(x)$ and I think you can do the rest.