# Problem Of The Week # 343 - Mar 22, 2019

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#### Euge

##### MHB Global Moderator
Staff member
Here is this week's POTW:

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Evaluate the sum of the series

$$\sum_{n = 0}^\infty \frac{(-1)^n}{2n+1} \sech\left[\frac{(2n+1)\pi}{2}\right]$$

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If $S$ is the sum, then $$S = \sum_{n = -\infty}^0 \dfrac{(-1)^n}{-2n+1} \sech\left[\frac{(-2n+1)\pi}{2}\right] = \sum_{n = -\infty}^{-1} \frac{(-1)^n}{2n+1} \sech\left[\frac{(2n+1)\pi}{2}\right]$$ so that $$2S = \sum_{n =-\infty}^\infty \frac{(-1)^n}{2n+1}\sech\left[\frac{(2n+1)\pi}{2}\right] = \frac{1}{2}\sum_{n = -\infty}^\infty \frac{(-1)^n}{n + \frac{1}{2}}\sech\left[\left(n + \frac{1}{2}\right)\pi\right]$$ The function $f(z) = \dfrac{\sech \pi z}{z}$ has $z f(z) \to 0$ as $\lvert z\rvert \to \infty$, so $2S$ is equal to one-half the sum of the residues of $f(z)\pi\sec(\pi z)$ at the singularities of $f$. Now $f$ has simple poles at $z = 0$ and at $z = -i\dfrac{2n+1}{2}$ where $n$ ranges over the integers. The residues of $f(z)\pi \sec(\pi z)$ at $0$ and $-i \dfrac{2n+1}{2}$, respectively, are $\pi$ and $-\dfrac{(-1)^n}{n + \frac{1}{2}}\sech\left[\left(n + \frac{1}{2}\right)\pi\right]$. Therefore, $2S = \dfrac{\pi}{2} - 2S$, or $S = \dfrac{\pi}{8}$.