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Using the Maclaurin series for $\sin z$ and $\sinh z$, as well as the infinite products $$\sin z = z\prod_{n=1}^\infty\left(1 - \frac{z^2}{n^2\pi^2}\right)$$ and $$\sinh z = z\prod_{n=1}^\infty\left(1 + \frac{z^2}{n^2\pi^2}\right)$$ deduce that $$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}$$

I am honestly not even sure where to begin for this problem. I have been staring at it for a while and I don't know how the Maclaurin series or infinite products of $\sin z$ and $\sinh z$ will help me evaluate the summation. Any hints to help me get on the right track would be greatly appreciated!

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If we start with: $$ \frac{\sin z}{z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right)\tag{1} $$ and take the logarithmic derivative of both sides we get: $$ -\frac{1}{z}+\cot(z) = \sum_{n\geq 1}\frac{2z}{z^2-n^2 \pi^2}\tag{2} $$ so, by expanding the general term of the RHS as a geometric series: $$ 1-z\cot z = 2\sum_{n\geq 1}\sum_{k\geq 1}\frac{z^{2k}}{n^{2k}\pi^{2k}} \tag{3}$$ then switching the sums and exploiting the definition of the $\zeta$ function: $$ 1-z\cot z = 2\sum_{k\geq 1}\frac{\zeta(2k)}{\pi^{2k}}z^{2k}\tag{4} $$ so the values of the Riemann $\zeta$ function at the even integers can be recovered from the Taylor series of $1-z\cot z$ at $z=0$. Since $$ 1-z\cot z = \frac{z^2}{3}+\frac{z^4}{45}+o(z^5) \tag{5}$$ it follows that:

$$ \zeta(4) = \frac{\pi^4}{2}\cdot\frac{1}{45} = \color{red}{\frac{\pi^4}{90}}\tag{6}$$

as wanted. An alternative approach comes from: $$ \frac{\sin(z)\sinh(z)}{z^2}=\prod_{n\geq 1}\left(1-\frac{z^4}{n^4 \pi^4}\right) \tag{7} $$ that directly leads to: $$ \zeta(4) = -\pi^4 [z^4]\frac{\sin(z)\sinh(z)}{z^2} = -\pi^4 [z^6] \sin(z)\sinh(z) \tag{8}$$ with the same outcome of $(6)$.

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