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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?

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    $\begingroup$ Your last step is wrong. You should get $B(1+1/2+...+1/A)+A(1+1/2+...+1/B)$. $\endgroup$ – Kavi Rama Murthy Apr 24 '19 at 12:15
  • $\begingroup$ how can B and A be in the multiple of series? $\endgroup$ – Pankaj Solanki Apr 24 '19 at 12:23
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    $\begingroup$ 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 $B$. $\endgroup$ – mihaild Apr 24 '19 at 12:35
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$$\sum_{n=1}^{A}{\left(\frac1n+\frac1m\right)}=\sum_{n=1}^{A}{\frac1n}+\sum_{n=1}^{A}{\frac1m}=H_A+\frac Am$$ where $H_A=\frac11+\frac12+...+\frac1A$ Thus:

$$\sum_{m=1}^B\sum_{n=1}^A \left(\frac{1}{n}+\frac{1}{m}\right)=\sum_{m=1}^{B}{\left(H_A+\frac Am\right)}$$ $$= {\sum_{m=1}^{B}{H_A}}+A\sum_{m=1}^{B}{\frac 1m}$$ $$=BH_A+AH_B$$

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