Question

Evaluate the sum:
\[\sum_{n=1}^\infty \arctan\frac{1}{n^2}\]

Answer 1

\[∑n =\infty \tan -1 (2n)\]

Answer 2

Did I do it correctly?

Answer 3

Anyway, the series is known to converge, since
\[\sum_{n=1}^\infty\arctan\frac{1}{n^2}<\int_0^\infty \arctan\frac{1}{x^2}\,dx\]
the evaluation of which is itself a fun exercise.

Answer 4

Learning this in the beginning myself

Answer 5

This is an infinite sum

Answer 6

Whoops, I meant
\[\int_{\color{red}1}^\infty\arctan\frac{1}{x^2}\,dx\]

Answer 7

nice problem

Answer 8

seems it requires some advanced methods, so far I think we should use the expansion of hyperbolic functions. There's some rules related to the summation to turn into pi not sigma, I think I need little refreshment in there. But for sure, will not be conventional series method

Answer 9

A solution by power series manipulation would definitely be a great thing to see.
\[\arctan x=\sum_{k=1}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}\]
As a matter of fact, the exact value of the series is in terms of the \(\arctan\) of an expression containing hyperbolic as well as trigonometric terms. The trick would be getting the rest to match up.

Answer 10

Well, by then you will have to find double summation here, for me I don't even know how to evaluate the first one.
The way you are going I THINK will stuck at a died end, or it maybe very tedious

Answer 11

One of the more interesting approaches I've come across is setting the sum to be the argument of an infinite product of complex numbers:
\[N=\prod_{k=1}^\infty (a_k+ib_k)~~\implies~~\arg N=\sum_{k=1}^\infty \arctan\frac{b_k}{a_k}\]
I'm still trying to wrap my mind around the steps beyond that, but I thought this initial reasoning was really clever!

Answer 12

I think this does rest of the job
http://www.math.umn.edu/~garrett/m/complex/hadamard_products.pdf

Answer 13

letting \(a_n = 1\) and \(b_n= \frac{1}{n^2}\)

\[\sum_{n=1}^\infty \arctan\frac{1}{n^2} = \arg \prod\limits_{k=1}^{\infty}(1+\frac{i}{n^2}) = \arg \dfrac{\sin \pi e^{i3\pi/4}}{\pi e^{i3\pi/4}} = \href{http:///www.wolframalpha.com/input/?i=arg++%5Cdfrac%7B%5Csin+%28%5Cpi+e%5E%7Bi3%5Cpi%2F4%7D%29%7D%7B%5Cpi+e%5E%7Bi3%5Cpi%2F4%7D%7D}{1.42\ldots}\]