In this thread, Jack D'Aurizio provided a succinct proof for the formula of calculating the values of the Zeta function $\zeta(2n)$
$$ \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}.\tag{2}$$$$\begin{eqnarray*} \coth z -\frac{1}{z} &=& \color{red}{\sum_{n\geq 1}}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right)\\&=&\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}\\&=&\color{red}{\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2\color{red}{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}\color{red}{n^{2m}}} \\&=&\sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}\tag{3}\end{eqnarray*}$$
and we have the claim by comparing the coefficients in the RHSs of (2) and (3).
I just have a couple of questions regarding this derivation.
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First, the sigma notation. I thought that the infinite product for $\sinh(x)$ is $\displaystyle\prod_{n=1}^{+\infty}\color{red}{z}\left(1+\dfrac{z^2}{n^2\pi^2}\right)$. So why there is a sum here, shouldn't it be a product? Also where does the $\color{red}z$ go?
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How to prove, or how do we know that $\coth(z)-\dfrac{1}{z}=\displaystyle\sum_{n=1}^{+\infty}\dfrac{d}{dz}\log\left(1+\dfrac{z^2}{n^2\pi^2}\right)$. This is my guess. The derivative of $\log(\sinh(z))$ is $\coth(z)$. So that is why it is, am I correct?
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I don't understand anything at all about the double sigma notation. I am always in fear of double sigma notation because I haven't learnt how to manipulate them well. So what was Jack doing here? Suddenly there are $(-1)^{2m-1}$, $z^{2m-1}$, $\pi^{2m}$
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I don't understand anything when he says "comparing the coefficients in the RHSs of (2) and (3)." What should I suppose to do to understand this?
Best Answer
$$\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \implies \log(\sinh{z}) = \log(z)+ \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \\ \implies \log(\sinh{z})-\log(z) = \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) = \sum_{n=1}^\infty\log\left(1+\frac{z^2}{n^2\pi^2}\right)$$
Differentiating both sides we get $$\coth z -\frac{1}{z} = {\sum_{n\geq 1}}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right) = \sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}$$
Next, we write the summand as an infinite series. Since $\displaystyle \frac{1}{1+t} = \sum_{m=1}^{\infty}(-1)^{m-1}t^{m-1}$ we have $$\frac{2z}{\pi^2 n^2+z^2} = \frac{2z}{\pi^2 n^2} \cdot \frac{1}{1+\left(\frac{z}{\pi n}\right)^2} = \frac{2z}{\pi^2 n^2} \cdot\sum_{m=1}^{\infty}(-1)^{m-1} \left(\frac{z}{\pi n}\right)^{2m-2}$$
Which is equal to $\displaystyle \frac{2z}{\pi^2 n^2} \cdot\sum_{m=1}^{\infty}(-1)^{m-1} \frac{z^{2m-2}}{\pi^{2m-2}n^{2m-2}} = \sum_{m=1}^{\infty}(-1)^{m-1} \frac{z^{2m-1}}{\pi^{2m}n^{2m}}.$
So we have $$\coth z -\frac{1}{z} ={\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} $$
And summing over $n$ first we have $${\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} = {\sum_{m\geq 1}\sum_{n\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} ={\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}} \bigg(\sum_{n \ge 1} \frac{1}{n^{2m}}\bigg)$$
Since $$\displaystyle \sum_{n \ge 1} \frac{1}{n^{2m}}= \zeta(2m)$$
We have
$$\displaystyle\coth z -\frac{1}{z} = \sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}.$$
So, on the one hand
$$\displaystyle \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}$$ but on the other $$\displaystyle \coth z-\frac{1}{z} = \displaystyle \sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}$$
So $$\sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1} = \displaystyle \sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}$$
Reindexing the RHS with $n$
$$\sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1} = \displaystyle \sum_{n\geq 1}\frac{2\,\zeta(2n)}{\pi^{2n}}(-1)^{n-1}z^{2n-1}$$
Then comparing the coefficient of $z^{2n-1}$ on both sides
$$\frac{4^n\,B_{2n}}{(2n)!} =\frac{2\,\zeta(2n)}{\pi^{2n}}(-1)^{n-1} $$
And finally solving for $\zeta(2n)$ gives:
$$\zeta(2n) = \frac{4^n\,B_{2n} (-1)^{n-1}\pi^{2n}}{2(2n)! } =\frac{\,B_{2n} (-1)^{n-1}(2\pi)^{2n}}{2(2n)! }$$