"Evaluate:∑n=1∞arctan⁡(18n2)" was answered by our community

SnuluddidaPalbx4

Answered question

2022-02-24

Evaluate:

\sum_{n = 1}^{\infty} \arctan (\frac{1}{8 n^{2}})

marinacabellok6k

Beginner2022-02-25Added 4 answers

We can compute this sum by considering it as the argument of an infinite product. Here the argument of the nth factor of the product is

- \arctan (\frac{1}{8} n^{2})

\sum_{n = 1}^{\infty} (\frac{1}{8 n^{2}}) = - arg \prod_{n = 1}^{\infty} (1 - \frac{i}{8 n^{2}})

= - arg \prod_{n = 1}^{\infty} (1 - \frac{{(\sqrt{\frac{i}{8}})}^{2}}{n^{2}})

= - arg \frac{\sin (π \sqrt{\frac{i}{8}})}{π \sqrt{\frac{i}{8}}}

= - arg (\frac{(1 - i) \sqrt{2} cosh (\frac{π}{4})}{π} + \frac{(1 + i) \sqrt{2} sinh (\frac{π}{4})}{π})

= - \arctan (\frac{sinh (\frac{π}{4}) - cosh (\frac{π}{4})}{sinh (\frac{π}{4}) + cosh (\frac{π}{4})})

= a r o t (e^{\frac{π}{2}})

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?