Explained: "Find the sum: S=∑k=1∞1k2−a2 where a∈(0,1)"

Daniell Phillips

Answered question

2022-01-18

Find the sum:

S = \sum_{k = 1}^{\infty} \frac{1}{k^{2} - a^{2}}

where

a \in (0, 1)

sirpsta3u

Beginner2022-01-18Added 42 answers

Starting with the infinite product

\frac{\sin π x}{π x} = \prod_{k = 1}^{\infty} (1 - \frac{x^{2}}{k^{2}})

taking the logarithm of both sides gives

\log (\frac{\sin π x}{π x}) = \log (\prod_{k = 1}^{\infty} (1 - \frac{x^{2}}{k^{2}})) = \sum_{k = 1}^{\infty} \log (1 - \frac{x^{2}}{k^{2}})

Differentiation gives

\frac{π x}{\sin π x} (\frac{\cos π x}{x} - \frac{\sin π x}{π x^{2}}) = \sum_{k = 1}^{\infty} \frac{- 2 x}{k^{2} (1 - \frac{x^{2}}{k^{2}})}

which simplifies to

π \cot π x - \frac{1}{x} = - 2 x \sum_{k = 1}^{\infty} \frac{1}{k^{2} - x^{2}}

\sum_{k = 1}^{\infty} \frac{1}{k^{2} - x^{2}} = \frac{1}{2 x^{2}} - \frac{π \cot π x}{2 x}

Ethan Sanders

Beginner2022-01-19Added 35 answers

S \equiv \sum_{k = 1}^{\infty} \frac{1}{k^{2} - a^{2}} = \sum_{k = 0}^{\infty} \frac{1}{(k + 1 - a) (k + 1 + a)}

= \frac{ψ (1 - a) - ψ (1 + a)}{- 2 a} = \frac{ψ (1 - a) - [ψ (a) + \frac{1}{a}]}{- 2 a}

= \frac{1}{- 2 a} [π \cot (π a) - \frac{1}{a}]

= \frac{1}{2 a} [\frac{1}{a} - π \cot (π a)]

alenahelenash

Expert2022-01-24Added 556 answers

Super i like it

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