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Express $$\sum_{j=1}^{n}\sum_{i=1}^{n} \frac{1}{i(i+j)}$$ in terms of the harmonic numbers $H_n$.

I guess that there could be several approaches for doing this.

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Let $S$ be the desired sum. By interchanging the roles of $i$ and $j$, we have $$2S=\sum_i \sum_j \left( \frac{1}{i(i+j)}+\frac{1}{j(i+j)} \right) = \sum_i \sum_j \frac{1}{ij} =H_n^2.$$ Hence $S=H_n^2/2$.

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  • $\begingroup$ it will be better to write the answer as$\frac{1}{2}(H_n)^2..$ $\endgroup$ – Dr Zafar Ahmed DSc Jul 16 '19 at 7:51
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$\begin{array}\\ S &=\sum_{j=1}^{n}\sum_{i=1}^{n} \frac{1}{i(i+j)}\\ &=\sum_{j=1}^{n}\sum_{i=1}^{n} \frac1{j}(\frac1{i}-\frac1{i+j})\\ &=\sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{n} (\frac1{i}-\frac1{i+j})\\ &=\sum_{j=1}^{n}\frac1{j}\left(\sum_{i=1}^{n} \frac1{i}-\sum_{i=1}^{n}\frac1{i+j}\right)\\ &=\sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{n} \frac1{i}-\sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{n}\frac1{i+j}\\ &=H_n^2-\sum_{i=1}^{n}\sum_{j=1}^{n}\frac1{j}\frac1{i+j}\\ &=H_n^2-S\\ \text{so}\\ S &=\frac12 H_n^2\\ \end{array} $

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  • $\begingroup$ @ it will be better to write the answer as $\frac{1}{2}(H_n)^2..$ $\endgroup$ – Dr Zafar Ahmed DSc Jul 16 '19 at 7:52
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    $\begingroup$ You say either and I say potato. $\endgroup$ – marty cohen Jul 16 '19 at 12:50

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