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

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.

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 $$S=H_n^2/2$$.
• it will be better to write the answer as$\frac{1}{2}(H_n)^2..$ – Z Ahmed Jul 16 '19 at 7:51
$$\begin{array}\\ S &=\sum_{j=1}^{n}\sum_{i=1}^{n} \frac{1}{i(i+j)}\\ &=\sum_{j=1}^{n}\sum_{i=1}^{n} \frac1{j}(\frac1{i}-\frac1{i+j})\\ &=\sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{n} (\frac1{i}-\frac1{i+j})\\ &=\sum_{j=1}^{n}\frac1{j}\left(\sum_{i=1}^{n} \frac1{i}-\sum_{i=1}^{n}\frac1{i+j}\right)\\ &=\sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{n} \frac1{i}-\sum_{j=1}^{n}\frac1{j}\sum_{i=1}^{n}\frac1{i+j}\\ &=H_n^2-\sum_{i=1}^{n}\sum_{j=1}^{n}\frac1{j}\frac1{i+j}\\ &=H_n^2-S\\ \text{so}\\ S &=\frac12 H_n^2\\ \end{array}$$
• @ it will be better to write the answer as $\frac{1}{2}(H_n)^2..$ – Z Ahmed Jul 16 '19 at 7:52