# Thread: Infinite Series: Sum of inverse quartics

1. ## Infinite Series: Sum of inverse quartics

Does anyone know what $\sum_{n=1}^{\infty} \frac 1 {n^4}$ is? I think that it's $\frac{\pi^4} {90}$.

2. Originally Posted by Kep
Does anyone know what $\sum_{n=1}^{\infty} \frac 1 {n^4}$ is? I think that it's $\frac{\pi^4} {90}$.
correct.

3. Originally Posted by Kep
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.

4. 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.

5. Oh and since when was it an inequality?

6. Originally Posted by Kep
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

7. 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.

8. Originally Posted by Kep
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

9. Originally Posted by Drexel28
there are formulas for the values for arbitrary powers.
I assume that these formulae come from doing the same thing but with $\frac {\cot(z)} {z^s}$

10. Originally Posted by Kep
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.

11. I believe there is a better way to get the equality than by integration.

12. Originally Posted by Shanks
I believe there is a better way to get the equality than by integration.
Agreed. Fourier series for one.

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

14. 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.

15. 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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