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anonymous

  • one year ago

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

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  1. misssunshinexxoxo
    • one year ago
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    \[∑n =\infty \tan -1 (2n)\]

  2. misssunshinexxoxo
    • one year ago
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    Did I do it correctly?

  3. anonymous
    • one year ago
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    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.

  4. misssunshinexxoxo
    • one year ago
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    Learning this in the beginning myself

  5. misssunshinexxoxo
    • one year ago
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    This is an infinite sum

  6. anonymous
    • one year ago
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    Whoops, I meant \[\int_{\color{red}1}^\infty\arctan\frac{1}{x^2}\,dx\]

  7. anonymous
    • one year ago
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    nice problem

  8. anonymous
    • one year ago
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    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

  9. anonymous
    • one year ago
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    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.

  10. anonymous
    • one year ago
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    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

  11. anonymous
    • one year ago
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    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!

  12. ganeshie8
    • one year ago
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    I think this does rest of the job http://www.math.umn.edu/~garrett/m/complex/hadamard_products.pdf

  13. ganeshie8
    • one year ago
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    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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