TITLE: How to prove $ \sum_{k=0}^n \frac{(-1)^{n+k}{n+k\choose n-k}}{2k+1}=\frac{-2\cos\left(\frac{2(n-1)\pi}{3}\right)}{2n+1}$
QUESTION [17 upvotes]: How to prove 
$$\sum_{k=0}^n \binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=-\frac{2}{2n+1}\,\cos\left(\frac{2(n-1)\pi}{3}\right)\;\text{?}$$
I have a proof by induction for it, but it isn't simple! I want to seek an interesting proof for it. (Sorry for my English study is very bad)
Can you help me re-open this question?
Thank you very much!

REPLY [15 votes]: My Solution
We have: \begin{align*}\sum_{k=0}^n\dfrac{(-1)^{n+k}}{2k+1}{n+k\choose n-k}&=\sum_{k=0}^n \dfrac{(-1)^{n+k}(n+k)!}{(2k+1)!(n-k)!} \\ &=\dfrac{1}{2n+1}\sum_{k=0}^n\dfrac{(-1)^{n+k}(n+k)!\left[(n+k+1)+(n-k)\right]}{(2k+1)!(n-k)!}\\&=\dfrac{1}{2n+1}\sum_{k=0}^n\left[(-1)^{n+k}{n+1+k\choose 2k+1} -(-1)^{n-1+k}{n+k\choose 2k+1}\right]\\ &=\dfrac{1}{2n+1}\left(S_n-S_{n-1}\right)\end{align*}
Where: \begin{align*}S_n&=\sum_{k=0}^n(-1)^{n+k}{n+1+k\choose 2k+1}=\sum_{k=0}^n(-1)^{n+k}{n+1+k\choose n-k}\\ &=\sum_{k=0}^n(-1)^k{2n+1-k\choose k}\qquad\text{(reverse index)}\end{align*}
So, easy proof $\quad S_n=-S_{n-1}-S_{n-2}\quad$ (reference here)
$\{S_n\}_{n\ge 0}\;:\;\{1,-1,0,1,-1,0,1,-1,0,...\}$
$\{S_n-S_{n-1}\}_{n\ge 1}\;:\;\{-2,1,1,-2,1,1,-2,1,1,...\}$
DONE! :D