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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.

  1. 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?

  2. 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?

  3. 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 $(-1)^{2m-1}$, $z^{2m-1}$, $\pi^{2m}$

  4. 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?

RobPratt
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James Warthington
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  • (1) I'm not sure which bit exactly you're talking about (2) yes; (3) try writing the summand as a geometric series; (4) compare the coefficient of $z^{n-1}$ in both expansions. – NoName Feb 06 '21 at 19:36
  • @NoName could you post a fuller details in an answer, I still don't understand anything. – James Warthington Feb 06 '21 at 19:41
  • @NoName: My question is why there is a sigma notation instead of the product notation. Perhaps I miss some details here. – James Warthington Feb 06 '21 at 19:51
  • Because log turns products into sums, i.e. $\log(a_1a_2 \cdots a_n) = \log(a_1)+\log(a_2) +\cdots + \log(a_n)$ – NoName Feb 06 '21 at 20:01
  • @NoName, ok I got it, so the term is going to be $\log(z)+\log(1+\frac{z^2}{\pi^2}) .etc$ Where is the $\log(z)$ in this case? – James Warthington Feb 06 '21 at 20:10
  • ad 4. $\sum_{n\geq 1}\frac{4^n,B_{2n}}{(2n)!}z^{2n-1}=\sum_{n\geq 1}\frac{2,\zeta(2n)}{\pi^{2n}}(-1)^{n-1}z^{2n-1}\color{red}{\Rightarrow} \frac{4^n,B_{2n}}{(2n)!}=\frac{2,\zeta(2n)}{\pi^{2n}}(-1)^{n-1}$ $\Rightarrow \zeta(2n)= \frac{4^n\cdot B_{2n}\cdot \pi^{2n}}{(2n)!\cdot 2}(-1)^{n-1}$. But I´m not sure if the first implication (red) is true. – callculus42 Feb 06 '21 at 20:18
  • $$\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \implies \log(\sinh{z}) = \log(z)+ \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) $$

    I think your confusion is coming from that in your formula for $\sinh{z}$ you have the $z$ inside the product.

     – NoName Feb 06 '21 at 20:23
  • @NoName, thanks man. But in the answer of Jack, there is no $\log(z)$. There should be one, shouldn't there? I think he differentiate it to obtain $\dfrac{1}{z}$, then substract from $\dfrac{1}{z}$. Not sure if I am correct here. – James Warthington Feb 06 '21 at 20:25
  • Well if you rewrite it as $$ \log(\sinh{z})-\log(z) = \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) = \sum_{n=1}^\infty\log\left(1+\frac{z^2}{n^2\pi^2}\right)$$

    Then differentiate both sides then $\log{z}$ becomes $\frac{1}{z}$ and the RHS becomes your first red sum.

     – NoName Feb 06 '21 at 20:28
  • @NoName, ok thanks man! Now the double sigma notation part really kills me. I am struggling to interpret it correctly. – James Warthington Feb 06 '21 at 20:31
  • The next step is using the geometric series $\displaystyle \frac{1}{1+t} = \sum_{m=1}^{\infty}(-1)^{m-1}t^{m-1}$ as we've $$ \frac{2z}{\pi^2 n^2+z^2} = \frac{2z}{\pi^2 n^2} \cdot \frac{1}{1+\left(\frac{z}{\pi n}\right)^2} = \frac{2z}{\pi^2 n^2} \cdot\sum_{m=1}^{\infty}(-1)^{m-1} \left(\frac{z}{\pi n}\right)^{2m-2}$$

    Which is equal to $\displaystyle \frac{2z}{\pi^2 n^2} \cdot\sum_{m=1}^{\infty}(-1)^{m-1} \frac{z^{2m-2}}{\pi^{2m-2}n^{2m-2}} = \sum_{m=1}^{\infty}(-1)^{m-1} \frac{z^{2m-1}}{\pi^{2m}n^{2m}}$

     – NoName Feb 06 '21 at 20:59
  • @NoName, yeah I am also fumbling with your idea: I am trying to rewrite

    $$\displaystyle\sum_{n=1}^{\infty}\dfrac{2}{\pi^2n^2+z^2}=\displaystyle2\sum_{n=1}^{\infty}\dfrac{1}{\pi^2n^2+z^2}$$

    Since $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{\pi^2n^2+z^2}$ can be rewritten as a geometric series, I have:

    $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{\pi^2n^2+z^2}$=geometric series formula

     – James Warthington Feb 06 '21 at 21:01
  • I'm not sure which part you're misunderstanding now.

    We've already shown that $$\displaystyle \frac{2z}{\pi^2 n^2+z^2} = \sum_{m=1}^{\infty}\color{blue}{2}(-1)^{m-1} \frac{z^{2m-1}}{\pi^{2m}n^{2m}}$$

    (The blue 2 is missing in my last comment, can't edit anymore). Therefore $$\displaystyle\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2} = \color{red}{\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2\color{red}{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}\color{red}{n^{2m}}}$$

    So we've already arrived at the double sum.

     – NoName Feb 06 '21 at 21:08
  • @NoName, thank you very much. I got the geometric part already. Now I am confused with the part where he says "and we have the claim by comparing the coefficients in the RHSs of (2) and (3).: $\displaystyle\sum_{n\geq 1}\sum_{m\geq 1}\frac{2(-1)^{m-1}z^{2m-1}}{\pi^{2m}n^{2m}}=\displaystyle\sum_{m\geq 1}\frac{2,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}= \sum_{n\geq 1}\frac{4^n,B_{2n}}{(2n)!}z^{2n-1}$. The equality is not so obvious to me. – James Warthington Feb 06 '21 at 21:13
  • I am substituting values into the expression to see if they are reduced to one another. – James Warthington Feb 06 '21 at 21:22
  • Sum over $n$ first

    $$ \color{red}{\sum_{n\geq 1}\sum_{m\geq 1}}\frac{2\color{red}{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}\color{red}{n^{2m}}} = {\sum_{m\geq 1}\sum_{n\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}{n^{2m}}} ={\sum_{m\geq 1}}\frac{2{(-1)^{m-1}z^{2m-1}}}{\pi^{2m}} \bigg(\sum_{n \ge 1} \frac{1}{n^{2m}}\bigg) $$

    Since $\displaystyle \sum_{n \ge 1} \frac{1}{n^{2m}}= \zeta(2m)$, the sum becomes $\displaystyle \sum_{m\geq 1}\frac{2,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}$ as required.

    For con. power series: $\sum a_n x^n = \sum b_n x^n \implies a_n = b_n $ for all $n \in \mathbb{N}$.

     – NoName Feb 06 '21 at 21:41
  • Ok I see, the sum $\sum_{n=1}^{+\infty} \dfrac{1}{n^{2m}}$ is indeed $\zeta{2m}$. It is that simple yet I keep banging my heads. Thank you so much. I wish I could upvote you but you didn't make a reply. The comments cannot be modified easily so I can't do anything for you. – James Warthington Feb 06 '21 at 21:58

1 Answers1

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$$\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \implies \log(\sinh{z}) = \log(z)+ \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) \\ \implies \log(\sinh{z})-\log(z) = \log\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right) = \sum_{n=1}^\infty\log\left(1+\frac{z^2}{n^2\pi^2}\right)$$

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:

$$\zeta(2n) = \frac{4^n\,B_{2n} (-1)^{n-1}\pi^{2n}}{2(2n)! } =\frac{\,B_{2n} (-1)^{n-1}(2\pi)^{2n}}{2(2n)! }$$

NoName
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