Simplifying Sum

You can prove it by induction on $n$. First observe that the right-hand side is

$$\frac1{n+m+1}\binom{n+m}n^{-1}\;,$$

and multiply through by $\binom{n+m}n$. Now suppose as induction hypothesis that

$$\sum_{i=0}^n\binom{n}i\binom{n+m}n(-1)^i\frac1{m+1+i}=\frac1{m+1+n}$$

for all $m\ge 0$. Then

$$\begin{align*} \sum_{i=0}^{n+1}&\binom{n+1}i\binom{n+1+m}{n+1}(-1)^i\frac1{m+1+i}\\ &=\sum_{i=0}^{n+1}\left(\binom{n}i+\binom{n}{i-1}\right)\binom{n+1+m}m(-1)^i\frac1{m+1+i}\\ &=\frac{m+1+n}{n+1}\sum_{i=0}^n\binom{n}i\binom{n+m}m(-1)^i\frac1{m+1+i}\\ &\qquad+\sum_{i=0}^n\binom{n}i\binom{n+1+m}{m+1}\cdot\frac{m+1}{n+1}\cdot(-1)^{i+1}\frac1{m+2+i}\\ &\overset{(*)}=\frac1{n+1}-\frac{m+1}{n+1}\cdot\frac1{m+2+n}\\ &=\frac1{n+1}\left(1-\frac{m+1}{m+2+n}\right)\\ &=\frac1{m+2+n}\;, \end{align*}$$

as desired, where the step $(*)$ applies the induction hypothesis both at $m$ and at $m+1$.