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 $x\ne 0$ .

#Calculus

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Comments

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 - 5 years, 1 month 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.

Mark Hennings - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$