Is there a way to simplify $\sum_{j=0}^{n}C_{n}^{j}\sum_{i=0}^{m}\frac{j}{j+i}C_{m}^{i}$ Is there a way to simplify the following sequence?
$$
\sum_{j=0}^{n}C_{n}^{j}\sum_{i=0}^{m}\frac{j}{j+i}C_{m}^{i}
$$

Recall that $$j{n\choose j}=n{n-1\choose j-1},\qquad \frac1k=\int_0^1x^{k-1}\mathrm dx,$$ for every $k\geqslant1$, hence the sum in the question is $$n\sum_{i,j}{n-1\choose j-1}{m\choose i}\int_0^1x^{j+i-1}\,\mathrm dx=n\int_0^1\sum_j{n-1\choose j-1}x^{j-1}\cdot\sum_i{m\choose i}x^i\,\mathrm dx,$$ that is, $$n\int_0^1(1+x)^{n-1}\cdot(1+x)^m\,\mathrm dx=\frac{n}{n+m}\left.(1+x)^{n+m}\right|_0^1=\frac{n}{n+m}(2^{n+m}-1).$$
