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Algebra Difficulty 5.2 AIME, harder Prove it Ireland

Prove that ∑m=1M1m(m+1)∑k=1m+11k<2\sum_{m=1}^{M} \frac{1}{m(m+1)} \sum_{k=1}^{m+1} \frac{1}{k} < 2.

Solution

Let Hn=∑k=1n1kH_n = \sum_{k=1}^{n} \frac{1}{k}, then Hm+1−Hmm=1m(m+1)=1m−1m+1\frac{H_{m+1} - H_m}{m} = \frac{1}{m(m+1)} = \frac{1}{m} - \frac{1}{m+1}, and
∑m=1M1m(m+1)∑k=1m+11k=∑m=1M1m(m+1)Hm+1=∑m=1M(1m−1m+1)Hm+1=∑m=1M1mHm+1−∑m=1M1m+1Hm+1=H2+∑m=2M1mHm+1−∑m=2M+11mHm=32+∑m=2M1m(Hm+1−Hm)−1M+1HM+1<32+∑m=2M(1m−1m+1)=2−1M+1<2. \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*}

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