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

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misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0\[∑n =\infty \tan 1 (2n)\]

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Did I do it correctly?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyway, 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.

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Learning this in the beginning myself

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0This is an infinite sum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, I meant \[\int_{\color{red}1}^\infty\arctan\frac{1}{x^2}\,dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0seems 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One 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!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2I think this does rest of the job http://www.math.umn.edu/~garrett/m/complex/hadamard_products.pdf

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2letting \(a_n = 1\) and \(b_n= \frac{1}{n^2}\) \[\sum_{n=1}^\infty \arctan\frac{1}{n^2} = \arg \prod\limits_{k=1}^{\infty}(1+\frac{i}{n^2}) = \arg \dfrac{\sin \pi e^{i3\pi/4}}{\pi e^{i3\pi/4}} = \href{http:///www.wolframalpha.com/input/?i=arg++%5Cdfrac%7B%5Csin+%28%5Cpi+e%5E%7Bi3%5Cpi%2F4%7D%29%7D%7B%5Cpi+e%5E%7Bi3%5Cpi%2F4%7D%7D}{1.42\ldots}\]
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