Evaluate $\sum\limits_{k=0}^n(-1)^{n+k}{n\choose k}{ {n+k}\choose n} \frac{1}{k+2}$ I am trying to evaluate the following sum :
$S=\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose n} \frac{1}{k+2}$
which is the same as 
$\displaystyle\sum_{k=0}^n(-1)^{n+k}{n\choose k} {{n+k}\choose k} \frac{1}{k+2}$
My approach so far has been aimed at converting this sum to a known form (given as a standard result in Integrals and series by Prudnikov et.al.)
$\displaystyle\sum_{k=1}^n(-1)^{k+1}{n\choose k} {{n+k}\choose k} \frac{1}{k}=\sum_{k=1}^n\frac{1}{k}$
I thought of combining the like terms ${n\choose k} {{n+k}\choose k}$ in the two series but that again depends on whether $n$ is even or odd. Any suggestion to proceed further ?

$$
\begin{align}
&\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{n}\frac1{k+2}\\[6pt]
&=\sum_{k=0}^n(-1)^{n+k}\binom{n}{k}\binom{n+k}{k}\left(\frac1{k+1}-\frac1{(k+1)(k+2)}\right)\tag{1}\\[6pt]
&=\frac1{n+1}\sum_{k=0}^n(-1)^{n+k}\binom{n+1}{k+1}\binom{n+k}{k}\\
&-\frac1{(n+1)(n+2)}\sum_{k=0}^n(-1)^{n+k}\binom{n+2}{k+2}\binom{n+k}{k}\tag{2}\\[6pt]
&=\frac{(-1)^n}{n+1}\sum_{k=0}^n\binom{n+1}{n-k}\binom{-n-1}{k}\\
&-\frac{(-1)^n}{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}{n-k}\binom{-n-1}{k}
\tag{3}\\[6pt]
&=\frac{(-1)^n}{n+1}\binom{0}{n}-\frac{(-1)^n}{(n+1)(n+2)}\binom{1}{n}\tag{4}\\[6pt]
&=\bbox[5px,border:2px solid #C0A000]{\left\{\begin{array}{}
\frac12&\text{if }n=0\\
\frac16&\text{if }n=1\\
0&\text{if }n\ge2
\end{array}\right.}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\frac1{k+2}=\frac1{k+1}-\frac1{(k+1)(k+2)}$ and $\binom{n+k}{n}=\binom{n+k}{k}$
$(2)$: $\frac1{k+1}\binom{n}{k}=\frac1{n+1}\binom{n+1}{k+1}$ applied once to the first sum and twice to the second
$(3)$: $\binom{n+j}{k+j}=\binom{n+j}{n-k}$ and $\binom{n+k}{k}=(-1)^k\binom{-n-1}{k}$
$(4)$: Vandermonde's Identity
$(5)$: evaluate at $n=0$ and $n=1$, and for $n\ge2$, $\binom{0}{n}=\binom{1}{n}=0$
