Jun 17, 2021 - 2:34

Problem

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— 級数bot (@infseriesbot) June 17, 2021

Solution

We have

$$ \begin{aligned} &\frac{1}{2^2-1} + \frac{1}{3^2-1} + \frac{1}{4^2-1} + \cdots \\
&= \frac{1}{2}\left( \frac{1}{2-1} - \frac{1}{2+1} + \frac{1}{3-1} - \frac{1}{3 + 1} + \cdots \right) \\
&= \frac{1}{2}\left( 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \cdots \right) \\ &= \frac{1}{2}\cdot\frac{3}{2} = \frac{3}{4} \end{aligned} $$

We have

$$ \begin{aligned} &\frac{1}{3^2-2^2} + \frac{1}{4^2-2^2} + \frac{1}{5^2-2^2} + \cdots \\
&= \frac{1}{4}\left(1 - \frac{1}{5} + \frac{1}{2} - \frac{1}{6} + \frac{1}{3} - \frac{1}{7} + \frac{1}{4} - \frac{1}{8} \cdots \right) \\
&= \frac{1}{4}(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = \frac{25}{48} \end{aligned} $$