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Prove that the sum from 1 to ∞ of 1 / (n2 – 1) = 3/4

Prove that the following sum converges and has the given value.

    \[ \sum_{n=1}^{\infty} \frac{1}{n^2 - 1} = \frac{3}{4}. \]


First, we need to get the terms in the sum into the form b_n - b_{n+1} so we can use the theorem on telescoping sums (Theorem 10.4 on page 386 of Apostol).

    \begin{align*}  \sum_{k=2}^{\infty} \frac{1}{n^2 - 1} &= \sum_{n=2}^{\infty} \frac{1}{(n+1)(n-1)} \\[9pt]  &= \frac{1}{2} \sum_{n=2}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n+1} \right) \\[9pt]  &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right)\\[9pt]  &= \frac{1}{2} \sum_{n=1}^{\infty} \left( \left( \frac{1}{n} + \frac{1}{n+1} \right) - \left( \frac{1}{n+1} + \frac{1}{n+2}\right) \right) \\[9pt]  &= \frac{1}{2} \sum_{n=1}^{\infty} (b_n - b_{n+1})  \end{align*}

where b_n = \frac{1}{n} + \frac{1}{n+1}. Hence,

    \begin{align*}  2 \sum_{n=2}^{\infty} \frac{1}{n^2-1} &= b_1 - L \\[9pt]  &= 1 + \frac{1}{2} - 0 \\[9pt]  &= \frac{3}{2}. \end{align*}

Hence,

    \[ \sum_{n=2}^{\infty} \frac{1}{n^2-1} = \frac{3}{4}. \]

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