n-1-H-n-n-n-1-pi-2-6-

Question Number 166254 by mnjuly1970 last updated on 16/Feb/22

Θ = \underset{n = 1}{\sum^{\infty}} \frac{H_{n}}{n . (n + 1)} \overset{?}{=} \frac{π^{2}}{6}

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Answered by Kamel_Ben last updated on 16/Feb/22

Θ = \underset{n = 1}{\sum^{+ \infty}} \frac{H_{n + 1} - \frac{1}{n + 1}}{n (n + 1)} = \underset{n = 1}{\sum^{+ \infty}} \frac{H_{n + 1}}{n (n + 1)} - \underset{n = 1}{\sum^{+ \infty}} \frac{1}{n (n + 1)} + \underset{n = 1}{\sum^{+ \infty}} \frac{1}{{(n + 1)}^{2}}

= \underset{n = 1}{\sum^{+ \infty}} (\frac{H_{n}}{n} - \frac{H_{n + 1}}{n + 1}) + \frac{π^{2}}{6} - 1 = \frac{π^{2}}{6}