# Riemann Zeta Function: What Is $ζ(2n)$?

Throughout this post, I will let $s:=2n$.
Recall from Calculus the Taylor Series for $\sin{x}$: $$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$ Now divide both sides by $x$ to find the Taylor Series for $\frac{\sin{x}}{x}$: $$\frac{\sin{x}}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots$$
Throughout this post, I will let $s:=2n$.
Recall from Calculus the Taylor Series for $\sin{x}$: $$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots$$ Now divide both sides by $x$ to find the Taylor Series for $\frac{\sin{x}}{x}$: $$\frac{\sin{x}}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots$$ Now, we need to find the zeroes of the sine function. Using trigonometry knowledge, recall that the zeroes occur when $x=k\pi\,\forall k\in\mathbb Z$. Therefore, we have the following factorization for $\frac{\sin{x}}{x}$: $$\frac{\sin{x}}{x}=\left(1-\frac{x}{\pi}\right)\cdot\left(1+\frac{x}{\pi}\right)\cdot\left(1-\frac{x}{2\pi}\right)\cdot\left(1+\frac{x}{2\pi}\right)\cdot\ldots=\left(1-\frac{x^2}{\pi^2}\right)\cdot\left(1-\frac{x^2}{2^2\pi^2}\right)\cdot\ldots$$ In a recent post I had proved this. This function, $\frac{\sin{x}}{x}$, is commonly referred to as Euler’s infinite product, and has a general formula: $$\frac{\sin{x}}{x} = \prod_{k=1}^{\infty} \left( 1 - \frac{x^2}{(k\pi)^{2}}\right) =\left(1-\frac{x^2}{\pi^2}\right)\cdot\left(1-\frac{x^2}{2^2\pi^2}\right)\cdot\ldots$$ Now, let us substitute $x=\pi s$: $$\implies \frac{\sin{\pi s}}{\pi s}=\left(1-s^2\right)\cdot\left(1-\frac{s^2}{2^2}\right)\cdot\ldots=\prod_{k=1}^{\infty}\left(1-\frac{s^2}{k^2}\right)$$ A simple trick we can use here, is to take the $\log$ of the product to make it a sum: $$\implies \log{\left(\frac{\sin{\pi s}}{\pi s}\right)}=\log{\left(\prod_{k=1}^{\infty}\left(1-\frac{s^2}{k^2}\right)\right)}=\sum_{k=1}^{\infty}\log{\left(1-\frac{s^2}{k^2}\right)}$$ $$\implies \log{\left(\sin{\pi s}\right)}=\log{\pi s}+\sum_{k=1}^{\infty}\log{\left(1-\frac{s^2}{k^2}\right)}$$ If we differentiate each side with respect to $s$, we will get: $$\pi\cdot\frac{\cos{\pi s}}{\sin{\pi s}}=\frac{1}{s}+\sum_{k=1}^{\infty}\frac{1}{\left({1-\frac{s^2}{k^2}}\right)}\cdot\left(-\frac{2s}{k^2}\right)$$ Now, multiply each side by $s$: $$\pi s\cdot\frac{\cos{\pi s}}{\sin{\pi s}}=1+\sum_{k=1}^{\infty}\frac{1}{\left({1-\frac{s^2}{k^2}}\right)}\cdot\left(-\frac{2s^2}{k^2}\right)$$ What do you notice on the LHS. You should notice that $\frac{\cos{\pi s}}{\sin{\pi s}}=\cot{\pi s}$. This implies: $$\pi s\cdot\frac{\cos{\pi s}}{\sin{\pi s}}=\pi s\cot{\pi s}$$ $$\implies \pi s\cot{\pi s}=1-2\displaystyle\sum_{k=1}^{\infty} \left[\left(1+\frac{s^2}{k^2}+\frac{s^4}{k^4}+\ldots\right)\left(\frac{s^2}{k^2}\right)\right]=1-2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \left(\frac{s^2}{k^2}\right)^n$$ $$=1-2\displaystyle\sum_{n=1}^{\infty}\left(\frac{s^{2n}}{1^{2n}}+\frac{s^{2n}}{2^{2n}}+\ldots\right)=1-2\sum_{n=1}^{\infty} \zeta (2n)\cdot s^{2n}$$ Can we find some other way to represent $\pi s\cot{\pi s}$? A little knowledge of complex analysis is required beyond this point. Notice that we could do the following: $$\pi s\cot{\pi s}=\pi s \cdot \frac{\cos{\pi s}}{\sin{\pi s}}=\pi s\cdot \frac{e^{i\pi s}+e^{-i\pi s}}{2}\cdot\frac{2i}{e^{i\pi s}-e^{-i\pi s}}=\pi s i\cdot \frac{e^{i\pi s}+e^{-i\pi s}}{e^{i\pi s}-e^{-i\pi s}}$$ When we multiply both the numerator and the denominator by $e^{i\pi s}$: $$\pi s\cot{\pi s}=\pi s i \cdot \frac{e^{2i\pi s}+1}{e^{2i\pi s}-1}=i\pi s \cdot\frac{2i\pi s}{e^{2i\pi s}-1}$$ Now, we have to find the infinite series for $\frac{z}{e^z-1}$. Let us assume the following is an infinite series of $\frac{z}{e^z-1}$: $$\frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{\beta_n z^n}{n!}$$ If we multiply each side by $e^z-1$, then by using the Taylor Series expansion for $\frac{e^z}{z}$: $$1=\sum_{n=0}^{\infty}\frac{\beta_n z^n}{n!}\sum_{n=0}^{\infty}\frac{z^n}{(n+1)!}$$ The Cauchy Product Formula states that: $$\left(\sum_{n=0}^{\infty}a_n\right)\left(\sum_{n=0}^{\infty}b_n\right)=\sum_{j=0}^{\infty}c_j, c_j=\sum_{k=0}^j{a_kb_{j-k}}$$ Therefore, $$1=\sum_{n=0}^{\infty}\frac{\beta_n z^n}{n!}\sum_{n=0}^{\infty}\frac{z^n}{(n+1)!}\implies 1=\sum_{j=0}^{\infty}\sum_{k=0}^j\frac{\beta_kz^j}{k!(j-k+1)!}$$ $$\implies 1=\displaystyle\sum_{j=0}^{\infty}\frac{1}{(j+1)!}\sum_{k=0}^{j} {\left(\begin{array}{c}j+1 \\k\end{array}\right)}\beta_kz^k$$ Since when $j=0$, $\beta=1$. This implies that the output is $1$. Therefore, the sum of all the other terms will be $0$. However this contradicts that fact that we set $z\ne 0$. Therefore, $$\displaystyle\sum_{k=0}^{j} {\left(\begin{array}{c}j+1 \\k\end{array}\right)}\beta_k=0$$ Since we know that $\beta_0=1$, we can find a realtion between all $\beta_k$, i.e. find all the terms of the sequence. This brings up another concept. These numbers are referred to as Bernoulli numbers. These numbers are of high importance in analysis and number theory. So now that we have found the infinite series of $\frac{z}{e^z-1}$, we can do the following: $$\pi s \cot{\pi s}=i\pi s +\frac{2i\pi s}{e^{2i\pi s}-1}=\pi i s + \sum_{n=0}^{\beta_n}\frac{\beta_n(2\pi i s)^n}{n!}$$ Since the cotangent function is an even function, this implies that the powers in the series expansion will all be even, i.e. $\beta_{2k+1}=0\,\forall k>0$. This gives us the following: $$\pi s \cot{\pi s}=\pi i s+\frac{\beta_0}{0!}+\frac{\beta_1(2\pi i s)}{1!}+\sum_{n=2}^{\infty}{\frac{\beta_n(2\pi i s)^n}{n!}}$$ Since we have $\beta_1=-0.5$: $$\pi s\cot{\pi s}=1+\sum_{n=1}^{\infty}\frac{\beta_{2n}(2\pi i s)^{2n}}{2n!}=1-2\sum_{n=1}^{\infty}-\frac{\beta_{2n}(2\pi i s)^{2n}}{2(2n!)}=1-2\sum_{n=1}^{\infty}\zeta(2n)s^{2n}$$ Therefore, $$-\frac{\beta_{2n}(2\pi i s)^{2n}}{2(2n!)}=\zeta(2n)s^{2n}\implies \zeta(2n)=(-1)^{n+1}\frac{\beta_{2n}2^{2n-1}\pi^{2n}}{2n!}$$