How is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$

Question

If $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$, then find the value of $\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots$

Firstly how is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$?

Secondly, I thought $$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}\cdots=\frac{S}{2}$$ But answer given is $\frac{\pi^4}{96}$. Whats the mistake in this?

Edit:

I found a way to get the answer. $$\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}\cdots+\frac{1}{2^4}\left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots\right)$$ $$S=S_1+\frac{1}{2^4}S$$

You tagged this as "algebra-precalculus". I think much more than this is needed to understand any proof I can think of this result. — DonAntonio, Mar 06 '16 at 05:35
I am a high school student. I took this question from my test paper. — Aditya Dev, Mar 06 '16 at 05:35
No, since that is your level. The question in yellow can be done, and my answer addresses it. The other question is what I can't do without results more advanced. — DonAntonio, Mar 06 '16 at 05:40
See here for proofs that the sum is $\pi^4/90$: http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90 — Hans Lundmark, Mar 06 '16 at 09:53
And here's a pretty elementary (almost high-scool level?) proof of $\sum k^{-2} = \pi^2/6$: http://math.stackexchange.com/a/8353/1242 — Hans Lundmark, Mar 06 '16 at 09:56

score 8 · Accepted Answer · answered Mar 06 '16 at 05:39

8

If you already know or take as given the first result, then

$$\frac{\pi^4}{90}=\sum_{n=1}^\infty\frac1{n^4}=\sum_{n=1}^\infty\frac1{(2n)^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{(2n-1)^4}\implies$$

$$\sum_{n=1}^\infty\frac1{(2n-1)^4}=\left(1-\frac1{16}\right)\frac{\pi^4}{90}=\frac{\pi^4}{96}$$

The first result though is way over the high school level, at least for me.

answered Mar 06 '16 at 05:39

DonAntonio

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Since it is above high school level, then what level do you estimate it is? – S.C.B. Mar 06 '16 at 05:43
@MXYMXY Thank you. At least first half of undergraduate level, since the easiest way I know of proving it is by means of Fourier series. Other methods may perhaps use some powers series of some specific functions. I'd love to see if there is an elementary way to reach that, but I doubt it. – DonAntonio Mar 06 '16 at 05:46
Since I am young, I don't know that much about the proof, but the one I have heard of( and at least partially understand) is the one using $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$, although this is contradictory, since I don't know what the above equation means anyway. – S.C.B. Mar 06 '16 at 05:50

score 1 · Answer 2 · answered Mar 06 '16 at 10:18

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Notice:

$$\sum_{n=a}^{m}\frac{b}{n^c}=b\left[\zeta(c,a)-\zeta(c,m+1)\right]$$
$$\sum_{n=a}^{\infty}\frac{b}{n^c}=\lim_{m\to\infty}\sum_{n=a}^{m}\frac{b}{n^c}=\lim_{m\to\infty}b\left[\zeta(c,a)-\zeta(c,m+1)\right]=b\zeta(c,a)\space\text{ when }b=0\vee\Re(c)>1$$

So, when we solve your question:

$$\sum_{n=1}^{\infty}\frac{1}{n^4}=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{1}{n^4}=\lim_{m\to\infty}\left[\zeta(4,1)-\zeta(4,m+1)\right]=\zeta(4,1)=\zeta(4)=\frac{\pi^4}{90}$$

answered Mar 06 '16 at 10:18

Jan Eerland

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I don't follow: how $;\zeta(4)=\frac{\pi^4}{90};$ follows without dealing with the series $;\sum\frac1{n^4};$ first? Or at least: are the calculations to deduce $;\zeta(4)=\frac{\pi^4}{90};$ easier or more basic than the ones done on the series in some way? – DonAntonio Mar 06 '16 at 11:00

How is $\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}\cdots=\frac{\pi^4}{90}$

2 Answers2