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Problem 1221

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Algebra Difficulty 5.2 Prove it Irish Mathematical Olympiad · Ireland

Prove that $\sum_{m=1}^{M} \frac{1}{m(m+1)} \sum_{k=1}^{m+1} \frac{1}{k} < 2$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $H_n = \sum_{k=1}^{n} \frac{1}{k}$ , then $\frac{H_{m+1} - H_m}{m} = \frac{1}{m(m+1)} = \frac{1}{m} - \frac{1}{m+1}$ , and
$\begin{align*} \sum_{m=1}^{M} \frac{1}{m(m+1)} \sum_{k=1}^{m+1} \frac{1}{k} &= \sum_{m=1}^{M} \frac{1}{m(m+1)} H_{m+1} \\ &= \sum_{m=1}^{M} \left( \frac{1}{m} - \frac{1}{m+1} \right) H_{m+1} \\ &= \sum_{m=1}^{M} \frac{1}{m} H_{m+1} - \sum_{m=1}^{M} \frac{1}{m+1} H_{m+1} \\ &= H_2 + \sum_{m=2}^{M} \frac{1}{m} H_{m+1} - \sum_{m=2}^{M+1} \frac{1}{m} H_m \\ &= \frac{3}{2} + \sum_{m=2}^{M} \frac{1}{m} (H_{m+1} - H_m) - \frac{1}{M+1} H_{M+1} \\ &< \frac{3}{2} + \sum_{m=2}^{M} \left( \frac{1}{m} - \frac{1}{m+1} \right) = 2 - \frac{1}{M+1} < 2. \end{align*}$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.