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Question Number 147737 by liberty last updated on 23/Jul/21

\sum_{n ⩾ 1} \frac{1}{n^{2}} {(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n})}^{2} = ?

Answered by ArielVyny last updated on 24/Jul/21

{(\sum_{n ⩾ 1} \frac{1}{n})}^{2} ⩽ Σ \frac{1}{n^{2}} = \frac{π^{2}}{6}

{(\sum_{n ⩾ 1} \frac{1}{n})}^{2} \sum_{n ⩾ 1} \frac{1}{n^{2}} ⩽ \frac{π^{4}}{36}