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I am asked to "use the calculus of residues" to prove that $$\displaystyle\sum\limits_{n=1}^{\infty} \frac{1}{n^4}=\frac{\pi^4}{90}$$

I think I can do this given the Laurent series for $\cot z$ centered at the origin, but I don't know how to find the first few terms of the Laurent series (I can use Cauchy's Integral Formula to find the first coefficient).

Zubin Mukerjee
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  • See this answer: http://math.stackexchange.com/questions/384780/closed-form-for-sum-n-infty-infty-frac1n4a4/384839#384839 – Ron Gordon Nov 19 '14 at 06:23
  • It will be $\frac{1}{z}\frac{\cos z}{(\sin z)/z}$. Expand the last fraction by dividing. – Mike Nov 19 '14 at 06:24

2 Answers2

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The calculus of residues solution comes from integrating $\frac{\pi\cot{\pi z}}{z^4}$ counterclockwise around the appropriate curve. The solution is worked out here.

Community
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Stella Biderman
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We have $$\dfrac{\sin(x)}x = \prod_{n=1}^{\infty}\left(1-\dfrac{x^2}{n^2 \pi^2}\right)$$ The Taylor series for $$\dfrac{\sin(x)}x = 1-\dfrac{x^2}{3!} + \dfrac{x^4}{5!} \mp$$ Comparing the coefficient of $x^4$, we obtain $$\sum_{\overset{m,n=1}{m< n}}^{\infty} \dfrac1{m^2n^2\pi^4} = \dfrac1{120}$$ $$\sum_{\overset{m,n=1}{m< n}}^{\infty}\dfrac1{m^2n^2} = \dfrac{\left(\displaystyle\sum_{m=1}^{\infty} \dfrac1{m^2} \right)^2 - \displaystyle\sum_{m=1}^{\infty} \dfrac1{m^4}}2$$ Making use of the fact that $$\displaystyle\sum_{m=1}^{\infty} \dfrac1{m^2} = \dfrac{\pi^2}{6}$$ we obtain $$\dfrac{\pi^4}{120} = \dfrac{\dfrac{\pi^4}{36}-\zeta(4)}2$$ Hence, $$\zeta(4) = \dfrac{\pi^4}{36}-\dfrac{\pi^4}{60} = \dfrac{\pi^4}{12}\left(\dfrac13-\dfrac15\right) = \dfrac{\pi^4}{90}$$

Adhvaitha
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    He was asked to "use the calculus of residues" to prove this. – Winther Nov 19 '14 at 06:35
  • @Winther Yes, "he was asked". But what he asked here was to prove that $\zeta(4) = \dfrac{\pi^4}{90}$. That is what his question says at least. – Adhvaitha Nov 19 '14 at 07:26