What options do I have for this series? No idea how to do it. ∑k=1∞cos2(k)4k2−1

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Nylah Church

2022-01-21

What options do I have for this series? No idea how to do it.

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

Ronald Alvarez

Beginner2022-01-22Added 11 answers

Express the sum as follows:

\sum_{k = 1}^{\infty} \frac{\cos^{2} k}{4 k^{2} - 1} = \frac{1}{2} \sum_{k = 1}^{\infty} \frac{1}{4 k^{2} - 1} + \frac{1}{2} \sum_{k = 1}^{\infty} \frac{\cos 2 k}{4 k^{2} - 1}

Now, the first sum on the RHS is equal to

\frac{1}{2}

. This may be shown using residue theory very easily:

\sum_{k = - \infty}^{\infty} \frac{1}{4 k^{2} - 1} = - R e s_{z = \pm \frac{1}{2}} \frac{π \cot π z}{4 z^{2} - 1} = 0

which means that

2 \sum_{k = 1}^{\infty} \frac{1}{4 k^{2} - 1} = 0

For the other sum, break into partial fractions and reorganize. You end up with

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

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

That last sum may be derived from the well-known sum

\sum_{k = 0}^{\infty} \frac{\sin (2 k + 1)}{2 k + 1} = \frac{π}{4}

Therefore, I get as the sum

\sum_{k = 1}^{\infty} \frac{\cos^{2} k}{4 k^{2} - 1} = \frac{1}{4} + \frac{1}{4} (\cos 2 - \frac{π}{4} \sin 1 + 2 \sin^{2} 1)

= \frac{1}{2} - \frac{π}{8} \sin 1

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