Key to "Sum of series: ∑n=1∞14n2−1"

Rylan Duncan

Answered question

2022-01-23

Sum of series:

\sum_{n = 1}^{\infty} \frac{1}{4 n^{2} - 1}

utgyrnr0

Beginner2022-01-24Added 11 answers

We want to compute it fastly and use the formula

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

where

x = \frac{1}{2}

because

\sum_{k = 1}^{\infty} \frac{1}{4 k^{2} - 1} = \frac{1}{4} \sum_{k = 1}^{\infty} \frac{1}{k^{2} - {(\frac{1}{2})}^{2}}

Here you may find more information about this precious way.

Amari Larsen

Beginner2022-01-25Added 10 answers

Hint: Partial Fraction decomposition:

\frac{1}{4 n^{2} - 1} = \frac{1}{(2 n - 1) (2 n + 1)} = \frac{1}{2} [\frac{1}{2 n - 1} - \frac{1}{2 n + 1}]

You must then compute the closed form of

\sum_{n = 1}^{k} [\frac{1}{2 n - 1} - \frac{1}{2 n + 1}]

Can you do that? Note that

\sum_{n = 1}^{k} \frac{1}{2 n - 1} = \frac{1}{1} + \frac{1}{3} + \dots + \frac{1}{2 k - 1}

= \frac{1}{2 \cdot 0 + 1} + \frac{1}{2 \cdot 1 + 1} + \dots + \frac{1}{2 (k_{1}) + 1}

= \sum_{n = 0}^{k - 1} \frac{1}{2 n + 1}

= \sum_{n = 1}^{k} \frac{1}{2 n + 1} + \frac{1}{2 \cdot 0 + 1} - \frac{1}{2 k + 1}

RizerMix

Expert2022-01-27Added 656 answers

Hint: Work on

S_{n} = \sum_{k = 1}^{n} \frac{1}{4 k^{2} - 1}

and take its limit when

n \to \infty

. Note that

\frac{1}{4 n^{2} - 1} = \frac{1}{2 (2 n - 1)} - \frac{1}{2 (2 n + 1)}

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