# 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
2 years, 2 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,

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

$$\displaystyle \implies \sum_{n=1}^{\infty} \dfrac{x}{n^2 -x^2} = \boxed{\dfrac{1}{2} \left( \dfrac{1}{x} - \dfrac{\pi x \cot (\pi x)}{x} \right)}$$

- 2 years, 2 months ago

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.

- 2 years, 1 month ago