anonymous
  • anonymous
Evaluate the sum: \[\sum_{n=1}^\infty \arctan\frac{1}{n^2}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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misssunshinexxoxo
  • misssunshinexxoxo
\[∑n =\infty \tan -1 (2n)\]
misssunshinexxoxo
  • misssunshinexxoxo
Did I do it correctly?
anonymous
  • anonymous
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.

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misssunshinexxoxo
  • misssunshinexxoxo
Learning this in the beginning myself
misssunshinexxoxo
  • misssunshinexxoxo
This is an infinite sum
anonymous
  • anonymous
Whoops, I meant \[\int_{\color{red}1}^\infty\arctan\frac{1}{x^2}\,dx\]
anonymous
  • anonymous
nice problem
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
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!
ganeshie8
  • ganeshie8
I think this does rest of the job http://www.math.umn.edu/~garrett/m/complex/hadamard_products.pdf
ganeshie8
  • ganeshie8
letting \(a_n = 1\) and \(b_n= \frac{1}{n^2}\) \[\sum_{n=1}^\infty \arctan\frac{1}{n^2} = \arg \prod\limits_{k=1}^{\infty}(1+\frac{i}{n^2}) = \arg \dfrac{\sin \pi e^{i3\pi/4}}{\pi e^{i3\pi/4}} = \href{http:///www.wolframalpha.com/input/?i=arg++%5Cdfrac%7B%5Csin+%28%5Cpi+e%5E%7Bi3%5Cpi%2F4%7D%29%7D%7B%5Cpi+e%5E%7Bi3%5Cpi%2F4%7D%7D}{1.42\ldots}\]

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