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Math Help - Infinite Series: Sum of inverse quartics

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    Kep
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    Infinite Series: Sum of inverse quartics

    Does anyone know what \sum_{n=1}^{\infty} \frac 1 {n^4} is? I think that it's \frac{\pi^4} {90}.
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    Quote Originally Posted by Kep View Post
    Does anyone know what \sum_{n=1}^{\infty} \frac 1 {n^4} is? I think that it's \frac{\pi^4} {90}.
    correct.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kep View Post
    Does anyone know what \sum_{n=1}^{\infty} \frac 1 {n^4} is? I think that it's \frac{\pi^4} {90}.
    Wolfram|Alpha. That inequality is very well known by the way.
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    Kep
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    Cool. I'd seen Euler's result about sum of inverse squares before, but this was new to me... glad to know that I've got it right.

    Yeah Wolfram Alpha is cool, but I couldn't work out what to query to get it to do it.
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    Kep
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    Oh and since when was it an inequality?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kep View Post
    Oh and since when was it an inequality?
    Haha, sorry I was just doing inequalities. So, how exactly did you work it out?

    P.S. there are formulas for the values for arbitrary powers.

    P.P.S. http://www.wolframalpha.com/input/?i...inity+of+1/n^4
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    Kep
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    By integrating \frac {\cot(z)} {z^4} around increasingly large squares (avoiding it's poles) using Cauchy's Residue Theorem and then showing that this integral goes to zero with the estimation lemma. The result follows. Coursework.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Kep View Post
    By integrating \frac {\cot(z)} {z^4} around increasingly large squares (avoiding it's poles) using Cauchy's Residue Theorem and then showing that this integral goes to zero with the estimation lemma. The result follows. Coursework.
    Ahh, I see haha
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    Kep
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    Quote Originally Posted by Drexel28 View Post
    there are formulas for the values for arbitrary powers.
    I assume that these formulae come from doing the same thing but with \frac {\cot(z)} {z^s}
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    Quote Originally Posted by Kep View Post
    By integrating \frac {\cot(z)} {z^4} around increasingly large squares (avoiding it's poles) using Cauchy's Residue Theorem and then showing that this integral goes to zero with the estimation lemma. The result follows. Coursework.
    I would be interested in seeing that if you feel like posting it. I know some CA, but I am certainly not what one would call well-versed in it. I would like to see that relationship. Sounds interesting.
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    Senior Member Shanks's Avatar
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    I believe there is a better way to get the equality than by integration.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Shanks View Post
    I believe there is a better way to get the equality than by integration.
    Agreed. Fourier series for one.
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    Kep
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    Wink

    I will post the method, but since it's assessed university coursework I will wait until the hand-in date has been and gone.
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    That's cool and understandable.

    I was thinking maybe you used \frac{cot({\pi}z)}{z^{4}}

    Did you use a rectangle centered at the origin with n+\frac{1}{2}, \;\ n-\frac{1}{2} along the real axis and -ni, \;\ ni along the imaginary axis?.

    I have seen it donw this way with \sum_{k=1}^{\infty}\frac{1}{k^{2}}

    The residue at 0 is
    \frac{-{\pi}^{3}}{45}

    and \frac{1}{{\pi}k^{4}} at k\neq 0

    I played around a little and combined the residues at -k and k into:

    2{\pi}i\cdot\frac{-{\pi}^{3}}{45}+4i\left(1+\frac{1}{2^{4}}+\frac{1}{  3^{4}}+....\right)=0


    \left(1+\frac{1}{2^{4}}+\frac{1}{3^{4}}+....\right  )=\frac{{\pi}^{4}}{90}

    Like I said, just playing around. CA is not my forte, but I enjoy it. It is a powerful tool. What I have learned is self-taught, so if there is a booboo I am here to learn as well as help.
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    Another way to find the general expression for \zeta(2n) is to expand \pi \cot \pi z in a Laurent series about 0 using the usual series expansion (whose coefficients are the Bernoulli numbers). Then expand it in another way using the identity \pi \cot \pi z = \frac{1}{z} + \sum_{j=1}^\infty\frac{1}{z^2-j^2} = \frac{1}{z} + \sum_{j=1}^\infty\frac{-1}{j^2}\frac{1}{1-z^2/j^2} = \frac{1}{z} + \sum_{j=1}^\infty\frac{-1}{j^2}\sum_{k=0}^\infty (z^2/j^2)^k. Switching the sums and gathering like powers, we obtain \zeta(2n)=\frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!} after comparing coefficients in both expansions.
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