### International Mathematics Competition for University Students

July 27 - Aug 2 2015, Blagoevgrad, Bulgaria

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### Problem 6

6. Prove that $$\sum\limits_{n = 1}^{\infty}\frac{1}{\sqrt{n}\left(n+1\right)} < 2.$$

Proposed by Ivan Krijan, University of Zagreb

Solution. We prove that $$\frac{1}{\sqrt{n}\left(n+1\right)} < \frac2{\sqrt{n}} - \frac2{\sqrt{n+1}}. \qquad\qquad (1)$$ Multiplying by $\sqrt{n}(n+1)$, the inequality (1) is equivalent with $$1 < 2(n+1) - 2\sqrt{n(n+1)}$$ $$2\sqrt{n(n+1)} < n + (n+1)$$ which is true by the AM-GM inequality.

Applying (1) to the terms in the left-hand side, $$\sum\limits_{n = 1}^{\infty} \frac{1}{\sqrt{n}\left(n+1\right)} < \sum\limits_{n = 1}^{\infty} \left( \frac2{\sqrt{n}} - \frac2{\sqrt{n+1}} \right) = 2.$$