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Some sum to sum

\[\large \sum _{n=1}^{\infty}\dfrac{x}{n^2-x^2}=\frac{1}{2}\left(\dfrac{1}{x}-{\pi \cot \left(\pi x\right)}\right)\]

Prove that the equation above holds true for \(x\ne 0 \).

Note by Julian Poon
9 months ago

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Consider the infinite product of \(\dfrac{\sin x}{x}\),

\( \displaystyle \dfrac{\sin x}{x} = \prod_{n=1}^{\infty} \left( 1 - \dfrac{x^2}{(n\pi)^{2}}\right) \)

Substitute \(x \mapsto \pi x\)

\(\implies \displaystyle \dfrac{\sin \pi x}{\pi x} = \prod_{n=1}^{\infty} \left( 1 - \dfrac{x^2}{n^{2}}\right)\)

Taking logarithm and differentiating w.r.t. \(x\) on both sides, we get,

\( \displaystyle - \sum_{n=1}^{\infty} \dfrac{2x}{n^2 -x^2} = \dfrac{\pi x \cot (\pi x)}{x} - \dfrac{1}{x}\)

\(\displaystyle \implies \sum_{n=1}^{\infty} \dfrac{x}{n^2 -x^2} = \boxed{\dfrac{1}{2} \left( \dfrac{1}{x} - \dfrac{\pi x \cot (\pi x)}{x} \right)} \) Ishan Singh · 9 months ago

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The function \[ F(z) \; = \; \frac{1}{z} + \sum_{n=1}^\infty \frac{2z}{z^2-n^2} \; = \; \sum_{n \in \mathbb{Z}} \frac{1}{z-n} \] is analytic on \(\mathbb{C} \backslash \mathbb{Z}\), periodic of period \(1\), and has simple poles with residue \(1\) at each integer. The same is true of the function \[ G(z) \; = \; \pi\cot \pi z \] Thus \(F(z) - G(z)\) is an entire function on \(\mathbb{C}\). It is not difficult to show that \(F-G\) is a bounded function - find a bound for \(|\mathrm{Re} z| \le \tfrac12\) and \(|\mathrm{Im} z| \ge 1\) by getting your hands dirty with the definition of each function, find a bound for \(|\mathrm{Re} z| \le \tfrac12\) and \(|\mathrm{Im} z| \le 1\) by continuity, and use periodicity to deduce boundedness everywhere.

Thus it follows that \(F-G\) is constant. Test the value at one point, such as \(z=\tfrac12\), to confirm that \(F\) and \(G\) are equal. Mark Hennings · 7 months, 3 weeks ago

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