Soit \(k\in\mathbb{N}^*\). Alors on a \[\begin{align*} S_k=& \sum_{n=1}^k \Big( \frac{1}{n}-\frac{3}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}\Big) \\ =& \sum_{n=1}^k \frac{1}{n} - 3 \sum_{n=1}^k \frac{1}{n+1}+\sum_{n=1}^k \frac{1}{n+2} + \sum_{n=1}^k \frac{1}{n+3} \\ =& \sum_{n=1}^k \frac{1}{n} - 3 \sum_{n=2}^{k+1} \frac{1}{n}+\sum_{n=3}^{k+2} \frac{1}{n} + \sum_{n=4}^{k+3} \frac{1}{n} \\ =& \Big( 1+\frac{1}{2}+\frac{1}{3}+ \sum_{n=4}^k \frac{1}{n}\Big) -3\Big(\frac{1}{2}+\frac{1}{3}+\sum_{n=4}^k \frac{1}{n} + \frac{1}{k+1} \Big) \\ &+\Big(\frac{1}{3}+\sum_{n=4}^k \frac{1}{n} + \frac{1}{k+1} + \frac{1}{k+2} \Big) +\Big(\sum_{n=4}^k \frac{1}{n}+ \frac{1}{k+1} + \frac{1}{k+2} +\frac{1}{k+3}\Big) \\ =& 1+\frac{1}{2}+\frac{1}{3}-\frac{3}{2}-\frac{3}{3}-\frac{3}{k+1} + \frac{1}{3}+ \frac{1}{k+1} + \frac{1}{k+2}+ \frac{1}{k+1} + \frac{1}{k+2} +\frac{1}{k+3} \end{align*}\] D’où pour tout \(k\in\mathbb{N}^*\), \[S_k=-\frac{1}{3} - \frac{1}{k+1} + \frac{2}{k+2}+\frac{1}{k+3}.\]

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