Expresar$\sum_{j=1}^{n}\sum_{i=1}^{n} \frac{1}{i(i+j)}$ en términos de números armónicos

Question

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

Supongo que podría haber varios enfoques para hacer esto.

Answer 1

Sea $S$ la suma deseada. Al intercambiar los roles de $i$ y $j$ , tenemos $$2S=\sum_i \sum_j \left( \frac{1}{i(i+j)}+\frac{1}{j(i+j)} \right) = \sum_i \sum_j \frac{1}{ij} =H_n^2.$$ Hence $ S = H_n ^ 2/2 $ .

Answer 2

$ \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} $