IMC 2015 · Day 2 · P6

We prove that $$\tag{1} \frac{1}{\sqrt{n(n+1)}} < \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}}.$$ Multiplying by $\sqrt{n(n+1)}$, the inequality (1) is equivalent with $$\begin{gather*} 1 < 2(n+1) - 2 \sqrt{n(n+1)} \\ 2 \sqrt{n(n+1)} < n + (n+1) \end{gather*}$$ which is true by the AM-GM inequality.

Applying (1) to the terms in the left-hand side, $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n(n+1)}} < \sum_{n=1}^{\infty} \left( \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{n+1}} \right) = 2.$$