Prove that, for positive integer $n$: $$ \sum_{k=0}^{n}\binom{n}{k} \binom{n+k+2}{k+1}^{-1}=\frac{1}{2}$$
Is a following correct:
We can rearrange the sum so that we have $$\sum_{k=0}^{n}(k+1)\binom{2n+2}{n+k+1} \binom{2n+2}{n+1}^{-1}=\frac{n+1}{2}$$
If we replace $k+1$ with $k'$ and $n+1$ with $n'$ then we have
$$\sum_{k'=0}^{n'}k'\binom{2n'}{n'+k'} \binom{2n'}{n'}^{-1}=\frac{n'}{2}$$
so we can write whole thing without $'$:
$$\sum_{k=0}^{n}k\binom{2n}{n-k} \binom{2n}{n}^{-1}=\frac{n}{2}$$
Now we prove this formula with some probabilistic arguments.
Say we have $n$ white and $n$ black balls and let us choose at random $n$ balls. Let $X$ count a number of white balls we choose.
Using indicator variable expression for $X$ we get $$E(X) = {n\over 2}$$ But clearly $E(X)$ is by the definition of the expectation the expression we want to count and we are done.
Your argument is fine, but there is a faster way: $$\begin{eqnarray*}\sum_{k=0}^{n}\binom{n}{k}\binom{n+k+2}{k+1}^{-1}&=&n!(n+1)!\sum_{k=0}^{n}\frac{k+1}{(n-k)!(n+k+2)!}\\&=&\frac{n!(n+1)!}{2}\sum_{k=0}^{n}\frac{(n+k+2)-(n-k)}{(n-k)!(n+k+2)!}\\&=&\frac{n!(n+1)!}{2}\sum_{k=0}^{n}\frac{1}{(n+k+1)!(n-k)!}-\frac{(n-k)}{(n-k)!(n+k+2)!}\end{eqnarray*}$$ the last sum is telescopic, and it equals $\frac{1}{(n+0+1)!(n-0)!}=\frac{1}{(n+1)!n!}$, proving the claim.