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

Question

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

My attempt

$$\sum_{k=0}^{n} \frac{(-1)^k}{2k+1}\binom{n}{k} = \\ \sum_{k=0}^{n} \frac{1}{2k+1}\binom{k-n-1}{k} = \\ \sum_{k=0}^{n} \frac{1}{2k+1}\binom{k-n-1}{-n-1} $$ now let $$ s:= -n - 1 = const $$

so $$\sum_{k=0}^{n} \frac{1}{2k+1}\binom{k-n-1}{-n-1} = \\ \sum_{k=0}^{n} \frac{1}{2k+1}\binom{k+s}{s} $$

And there I stucked. I was thinking if I can use that formula: $$ \sum_k \binom{k}{m} = \binom{n+1}{m+1} $$ but $ \frac{1}{2k+1}$ is an obstacle for me. How I can "remove" that?

Answer 1

$$(1+x^2)^n=\sum_{k=0}^n\binom nkx^{2k}$$

Integrate both sides wrt $x$

Set $x=0$ to find the constant of integration

Then $x^2=-1$