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Prove the sum from 1 to ∞ of 1 / ((2n-1)(2n+1)) = 1/2

by RoRi

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

    \[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}. \]

Denote the nth term in the sum by a_n, so we have

    \[ a_n = \frac{1}{(2n-1)(2n+1)} = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right). \]

Then, let

    \[ b_n = \frac{1}{2n-1} \quad \implies \quad b_{n+1} = \frac{1}{2n+1}. \]

Thus, 2a_n = b_n - b_{n+1}. Therefore, by Theorem 10.4 (page 386 of Apostol, on the convergence of sums of telescoping series) we know the series \sum a_n converges since the sequence \{ b_n \} converges. Furthermore, we can evaluate the sum,

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

Therefore,

    \[ \sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}. \]

One comment

  1. Abed says:

    Hi, thanks a lot for the work you have done, maybe I’m a little bit of a noob but before the end of the proof there is a 2 before the sum and I don’t know if its necessary

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