# Finding $\sum\limits_{m=1}^B\sum\limits_{n=1}^A\frac{m+n}{mn}$

Find$$\sum_{m=1}^B\sum_{n=1}^A \frac{m+n}{mn}.$$

What I tried:$$\sum_{m=1}^B\sum_{n=1}^A\frac{m+n}{mn}=\sum_{m=1}^B\sum_{n=1}^A\frac1n+\frac1m=\left(\frac11+\frac12+\cdots+\frac1A\right)+\left(\frac11+\frac12+\cdots+\frac1B\right).$$ This looks like harmonic progression and doesn't know how to sum. Now what will be the next step to get the answer?

• Your last step is wrong. You should get $B(1+1/2+...+1/A)+A(1+1/2+...+1/B)$. – Kavi Rama Murthy Apr 24 '19 at 12:15
• how can B and A be in the multiple of series? – Pankaj Solanki Apr 24 '19 at 12:23
• When you sum $\frac{1}{n}$ by $n$, you get the sum you wrote. But now you need to sum it by $m$. As terms are independent of $m$, you simply get term multiplied by number of terms, ie $B$. – mihaild Apr 24 '19 at 12:35

$$\sum_{n=1}^{A}{\left(\frac1n+\frac1m\right)}=\sum_{n=1}^{A}{\frac1n}+\sum_{n=1}^{A}{\frac1m}=H_A+\frac Am$$ where $$H_A=\frac11+\frac12+...+\frac1A$$ Thus:
$$\sum_{m=1}^B\sum_{n=1}^A \left(\frac{1}{n}+\frac{1}{m}\right)=\sum_{m=1}^{B}{\left(H_A+\frac Am\right)}$$ $$= {\sum_{m=1}^{B}{H_A}}+A\sum_{m=1}^{B}{\frac 1m}$$ $$=BH_A+AH_B$$