Can You Tackle This Advanced Series Summation Challenge?

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In summary, POTW stands for "Problem of the Week" and is a mathematical or scientific problem presented to the community to solve. The POTW is updated weekly and open to anyone with an interest in mathematics or science. While there are no monetary rewards, solving the POTW can improve problem-solving skills. Solutions are verified through community discussion and peer-review, and the problem poster may also provide their own solution.
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Euge
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MHB
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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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  • #2
No one answered this week's problem. You can read my solution below.

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 $S = \dfrac{\pi}{8}$.
 

FAQ: Can You Tackle This Advanced Series Summation Challenge?

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