<s>[INST] 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}$ 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!
 [/INST] Let's try a generating function approach.
Multiply by $x^ky^n$ and sum in both $k$ and $n$:
$$
\begin{align}
\sum_{k,n}\binom{n+k}{n-k}x^ky^n
&=\sum_{k,n}\binom{n}{2k}\frac{x^k}{y^k}y^n\tag{1}\\
&=\frac1{1-y}\sum_k\frac{x^ky^k}{(1-y)^{2k}}\tag{2}\\
&=\frac{1-y}{(1-y)^2-xy}\tag{3}
\end{align}
$$
Explanation:
$(1)$: $\binom{n+k}{n-k}=\binom{n+k}{2k}$, then substitute $n\mapsto n-k$
$(2)$: $\sum\limits_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$
$(3)$: $\sum\limits_kx^k=\frac1{1-x}$  
Substitute $x\mapsto-x^2$, $y\mapsto y^2$ and integrate $(3)$ in $x$ from $0$ to $1$:
$$
\begin{align}
\sum_{k,n}\frac{(-1)^k}{2k+1}\binom{n+k}{n-k}y^{2n}
&=\int_0^1\frac{1-y^2}{(1-y^2)^2+x^2y^2}\,\mathrm{d}x\tag{4}\\
&=\frac1y\,\tan^{-1}\!\left(\frac{y}{1-y^2}\right)\tag{5}\\[6pt]
&=\frac1y\left[\tan^{-1}\left(e^{i\pi/3}y\right)+\tan^{-1}\left(e^{-i\pi/3}y\right)\right]\tag{6}\\
&=2\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\cos\left(\frac{(2n+1)\pi}3\right)y^{2n}\tag{7}
\end{align}
$$
Explanation:
$(4)$: substitute $x\mapsto-x^2$, $y\mapsto y^2$ and integrate $(3)$ in $x$ from $0$ to $1$
$(5)$: perform the integration
$(6)$: use the formula for tangent of a sum
$(7)$: use the series for arctangent
Equation $(7)$ implies that
$$
\begin{align}
\sum_{k=0}^n\frac{(-1)^{n+k}}{2k+1}\binom{n+k}{n-k}
&=\frac2{2n+1}\cos\left(\frac{(2n+1)\pi}3\right)\tag{8}\\
&=-\frac2{2n+1}\cos\left(\frac{(2n-2)\pi}3\right)\tag{9}
\end{align}
$$
Explanation:
$(8)$: equate identical powers of $y$ in $(7)$
$(9)$: $\cos(x+\pi)=-\cos(x)$
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