what is the answer of
$$\sum_{m=1}^B\Pi_{n=1}^A \frac{m+n}{mn}=?$$
What I tried
$$\sum_{m=1}^B\Pi_{n=1}^A (\frac{1}{m} +\frac{1}{n})$$
$$\sum_{m=1}^B[(\frac{1}{m} +\frac{1}{1})(\frac{1}{m} +\frac{1}{2})...(\frac{1}{m} +\frac{1}{A})]$$
$$=[((\frac{1}{1} +\frac{1}{1})(\frac{1}{1} +\frac{1}{2})...(\frac{1}{1} +\frac{1}{A}))+((\frac{1}{2} +\frac{1}{1})(\frac{1}{2} +\frac{1}{2})...(\frac{1}{2} +\frac{1}{A}))+...+((\frac{1}{B} +\frac{1}{1})(\frac{1}{B} +\frac{1}{2})...(\frac{1}{B} +\frac{1}{A}))]$$
I get this answer. Can there be any further simplification or solution?