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I need to show that $$\sum_{n=1}^\infty{\frac{1}{n^8}} = \frac{\pi^8}{9450}$$ I have already shown that $$\sum_{n=1}^\infty{\frac{1}{n^4}} = \frac{\pi^4}{90}$$ by computing the Fourier series for the function $f(x)=x^2$ on the interval $(0,l)$ and then applying Parseval's Theorem. However, I don't know if $$\sum_{n=1}^\infty{\frac{1}{n^8}}$$ can be done in the same way because I am not given a function. My idea is to try and write $$\sum_{n=1}^\infty{\frac{1}{n^8}} = \sum_{n=1}^\infty({\frac{1}{n^4}})^2$$ so that I can use my previous calculations. Is this the right idea or is there an easier method? If this is the correct idea, then how does the square change the Fourier series?

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  • $\begingroup$ Try finding the Fourier coefficients of $f:[-\pi,\pi]\to\Bbb R$ defined by $f(t)=|t|^3$. $\endgroup$ – David C. Ullrich Sep 13 '18 at 17:03
  • $\begingroup$ @DavidC.Ullrich the second coefficient $b_n$ goes to 0. The coefficient $a_n = \frac{1}{\pi n^4}[(2\pi^3n^3-12\pi n)sin(\pi n)+6\pi^2n^2-12)cos(\pi n)+12]$ How do I simplify those? $\endgroup$ – mathqueen459 Sep 14 '18 at 4:55
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    $\begingroup$ That doesn't quite work, sorry. I started by looking for $f$ so that $f(t)=\sum n^{-3}e^{int}$, differentiating a few times to determine what $f$ should be. But of course that sum is nonsense, it should be $\sum_{n\ne0}$. If I fix what I had in mind it comes out to more or less the same argument as in the answer below, in reverse order. $\endgroup$ – David C. Ullrich Sep 14 '18 at 13:02
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Over the interval $(0,2\pi)$ we have $$ \frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n} \tag{1}$$ pointwise and uniformly over compact subsets. We may denote with $\text{Int}$ the operator which brings $f(x)$ into $C_f+\int_{0}^{x}f(t)\,dt$, where $C_f$ is chosen in such a way that $\int_{0}^{2\pi}(\text{Int}\,f)(x)\,dx = 0$. By applying such operator to both sides of $(1)$ we get $$ \frac{1}{12}\left(-2\pi^2+6\pi x-3x^2\right) = \sum_{n\geq 1}\frac{\cos(nx)}{n^2}\tag{2} $$ uniformly over any compact subset of $[0,2\pi]$, then by applying $-\text{Int}^2$ to both sides of $(2)$ we get $$ \sum_{n\geq 1}\frac{\cos(nx)}{n^4} = \text{Re}\,\text{Li}_4(e^{ix})=\frac{\pi^4}{90}-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac{1}{4}x^4.\tag{3} $$ Now we apply $\int_{0}^{2\pi}(\ldots)^2\,dx$ to both sides of $(3)$. We get:

$$ \pi \zeta(8) = \int_{0}^{2\pi}\left(\frac{\pi^4}{90}-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac{1}{4}x^4\right)^2\,dx=\begin{array}{c}\small\text{bunch of omitted}\\\small\text{computations}\end{array}=\frac{\pi^9}{9450}.\tag{4} $$ We are just integrating the square of a Bernoulli polynomial.


Anyway I believe that the most efficient approach for computing $\zeta(2m)$ for large values of $m\in\mathbb{N}^+$ is to recall Euler's generating function

$$ \frac{1-x\cot x}{2} = \sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}} x^{2m} \tag{4}$$ which comes from the application of $\frac{d}{dz}\log(\cdot)$ to both sides of $\frac{\sin z}{z}=\prod_{m\geq 1}\left(1-\frac{z^2}{\pi^2 m^2}\right)$.
Since $$\left[1-2\sum_{m\geq 1}\frac{\zeta(2m)}{\pi^{2m}}z^{2m}\right]\cdot\left[1+\sum_{m\geq 1}\frac{(-1)^m}{(2m+1)!}z^{2m}\right]=1+\sum_{m\geq 1}\frac{(-1)^m}{(2m)!}z^{2m}\tag{5}$$ we get (once $\zeta(0)$ is defined as $-\frac{1}{2}$, consistent with the analytic continuation of the series $\sum_{n\geq 1}\frac{1}{n^s}$) $$ \sum_{m=0}^{M}\frac{-2 \zeta(2m)(-1)^{M-m}}{\pi^{2m}(2M-2m+1)!}=\frac{(-1)^M}{(2M)!}\tag{6}$$ which allows to find $\zeta(2M)$ in terms of $\zeta(2M-2),\zeta(2M-4),\ldots,\zeta(2)$. This is equivalent to the well-known recursion for Bernoulli numbers.

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  • $\begingroup$ I'm not following what you are doing with the operator Int. How are we explicitly defining Int? $\endgroup$ – mathqueen459 Sep 14 '18 at 2:13
  • $\begingroup$ @mathqueen459: I wrote it: $$ (\text{Int}\,f)(x) = \int_{0}^{x}f(t)\,dt \underbrace{- \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{x}f(t)\,dt\,dx}_{C_f}$$ $\text{Int}$ picks an antiderivative, precisely the antiderivative with mean zero over $(0,2\pi)$. $\endgroup$ – Jack D'Aurizio Sep 14 '18 at 2:28
  • $\begingroup$ Ahh ok. Got it. Thank you! $\endgroup$ – mathqueen459 Sep 14 '18 at 2:46

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