Zeta function at even integers, deriving the formula for summing infinite series of the form $\sum_{n=1}^{+\infty}\frac{1}{n^{2m}}$

 My recent excursion to the general formula for the power series of trigonometric and hyperbolic $\cot(x)$ and $\coth(x)$ leads me further to one of the most beautiful and deep results in modern mathematics. The similarity between the closed form summing formula of the Zeta function at even integers and the general formula of $\cot(x)$ puzzles me, for I can see that there are some connections between the two. A long search on wikipedia and mathstackexchange proves my intuition. Yet, I am still a highly incompetent math student so I cannot, by myself, deduce the connection between $\cot(x)$ and the summing formula. The following derivation and results are owned not to me but to wikiproof and an answer of Jack D'Aurizio. Just compare between the general formula for $\cot(x)$ and the summing formula of Zeta function at even integers and you will see some connections:

$$\cot(x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

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

The similarity is striking to me, suggesting that there ought to be a connection of both of them. It turns out that my intuition is right. The relevant function is a modification of $\cot(x)$, which is the $\pi x\cot(\pi x)$. My initial guess is that the function $cot(\pi x)$ may be involved, but hours spent on worthless manipulation didn't lead me to the correct result. So as I mentioned above, I resorted to searching on google and I come up with $\pi x\cot(\pi x)$. 

History of problems

The history of the problems of summing infinite series of the form $\displaystyle\sum_{n=1}^{+\infty} \dfrac{1}{n^{2m}}$ is just as fascinating as the result itself. Infinite series had been a central focus of study of mathematicians in the XVII and XVIII century. Summing series was indeed very in vogue during this time period, and many, not just leading mathematicians, involved in this business for recreational purposes. The polymath Pierre Louis Maupertuis (1698-1759) made this following remark about this phenomenon in the XVIII century England:

"Cette affaire des suites qui est tout ce qu'il ly a de plus desagreable dans les mathematiques n'est qu'un jeu pour les Anglais, ce livre et celui de Moivre en sont une preuve."

"This business of series, the most disagreeable thing mathematics, is no more than a game for the English, this book and that of M.de Moivre are the proof."

Pietro Mengoli (1626-1686), an Italian mathematician who had studied with Bonaventura Cavalieri, in his "Novae quadraturae arithmeticae"  was successful in summing these telescoping series:

$$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n(n+2)}=\dfrac{1}{1\cdot 3}+\dfrac{1}{2\cdot 4}+\dfrac{1}{3\cdot5}...=\dfrac{3}{4}$$

$$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n(n+3)}=\dfrac{1}{1\cdot 4}+\dfrac{1}{2\cdot 5}+\dfrac{1}{3\cdot6}...=\dfrac{11}{18}$$

$$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot 3\cdot 4}+\dfrac{1}{3\cdot 4\cdot 5}...=\dfrac{1}{4}$$

$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{(2n+1)(2n+3)(2n+5)}=\dfrac{1}{3\cdot 5\cdot 7}+\dfrac{1}{5\cdot 7\cdot 9}+\dfrac{1}{7\cdot 9\cdot 11}...=\dfrac{1}{60}*$ *The book gives $\dfrac{1}{12}$, which is an error.

(See Ferraro, The rise and development of the theory of series up to the early 1820s, page 9).

He also successfully prove that the harmonic series is divergent $\sum_{n=1}^{+\infty}\dfrac{1}{n}$ (a result that was first proved by the late medieval French mathematician Nicole Oresme). More importantly, he shows that the alternating harmonic series converges to $\ln(2)$:

$$\displaystyle\sum_{n=1}^{+\infty}\frac{(-1)^n+1}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}...=\ln(2)$$

The now famous series:

$$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n^{2}}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}...=?$$

is beyond his power. He posed this problems to leading mathematicians of the time, yet it must be roughly 100 years later that Euler could finally provide various methods and proofs for the sum of this series, and similar series of higher even powers.

In his first textbook of teaching infinitesimal calculus to young students "Introductio a analysi infinitorum", he summed these series up to the 26th powers. The first 10 are included here:

$$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}...=\dfrac{\pi^2}{6}$$

$$1+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\dfrac{1}{4^4}+\dfrac{1}{5^4}+\dfrac{1}{6^4}...=\dfrac{\pi^4}{90}$$

$$1+\dfrac{1}{2^6}+\dfrac{1}{3^6}+\dfrac{1}{4^6}+\dfrac{1}{5^6}+\dfrac{1}{6^6}...=\dfrac{\pi^6}{945}$$

$$1+\dfrac{1}{2^8}+\dfrac{1}{3^3}+\dfrac{1}{4^8}+\dfrac{1}{5^8}+\dfrac{1}{6^8}...=\dfrac{\pi^8}{9450}$$

$$1+\dfrac{1}{2^{10}}+\dfrac{1}{3^{10}}+\dfrac{1}{4^{10}}+\dfrac{1}{5^{10}}+\dfrac{1}{6^{10}}...=\dfrac{\pi^{10}}{93555}$$

My "failed" attempt at deriving the generating function for $\zeta(2n)$

My failed attempt at deriving the generating function for $-\dfrac{\pi x}{2}\cot(\pi x)$

When I look at the formula, I know that I only have adjust some bits of the general formula of the $\cot(x)$ function to prove the summing formula for the value of $\zeta(2n)$. But I forgot just one small bit to complete my derivation. This is because I was quite tired after tinkering with the formula.

We have 

$$\cot(x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}(\pi x)^{2n-1}}{(2n)!}$$

$$-\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^{2n-1}}{(2n)!}$$

$$-x\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^\color{red}{{2n+1}}}{(2n)!}$$

$$-\dfrac{x}{2}\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^\color{red}{{2n+1}}}{2(2n)!}$$

I made the mistake red, it should be $2n$ and forgot to add another $\pi$ in to obtain the correct function. I thought I was wrong and running around looking for the right function. Wikiproof employs $\pi x\cot(\pi x)$ in its proof and I thought that my function was completely wrong. Only after seeing the generating function on wiki page of the Riemann zeta function that I know I was completely right, missing just one $\pi$.

So I have derived the formula all by myself, but  I was so unfortunate that I thought I was completely wrong. The correct function with its coefficients being $\zeta(2n)$ is:

$$-\dfrac{\pi x}{2}\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^{2n}}{(2n)!}=\sum_{n=0}^{+\infty}\zeta(2n)x^{2n}=\dfrac{1}{2}+\dfrac{\pi^2}{6}x^2+\dfrac{\pi^4}{90}x^4...$$

An interesting derivation of the formula of finding values of the Zeta function at even integers by Jack D'Aurizio

Jack D'Aurizio, an Italian mathematician on MathStackExchange provided this interesting derivation of the formula of finding values of the Zeta function at even integers. The starting point is not the $\cot(x)$ like my approach, but $\coth(x)$. I repost it here and add more intermediate steps to make it easier for readers to understand. The original answer is quite terse and omit middle steps: These middle steps are provided by NoName on Mathstack.

$$\frac{t}{e^t-1} = \sum_{n\geq 0}\frac{B_n}{n!}t^n \tag{1}$$

$$\coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}.\tag{2}$$

Jack expressed the function $\sinh(x)$ as the so-called Weierstrass product $\displaystyle\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)$ which was derived first by Euler.

Since:

$$\coth(z)=\dfrac{d}{dz}\log(\sinh(z))\tag{3}$$

$$\coth z-\frac{1}{z} = \dfrac{d}{dz}\displaystyle\sinh z=\dfrac{d}{dz}\log\left( z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)\right)$$

Using the law of log, we can rewrite

$$\log(z)+\log\left(\left(1+\dfrac{z^2}{\pi^2}\right)\left(1+\dfrac{z^2}{2\pi^2}\right)\left(1+\dfrac{z^2}{3\pi^2}\right)\right)...\tag{4}$$

$$\dfrac{1}{z}+\log\left(\left(1+\dfrac{z^2}{\pi^2}\right)\right)+\log\left(\left(1+\dfrac{z^2}{2\pi^2}\right)\right)+\log\left(\left(1+\dfrac{z^2}{3\pi^2}\right)\right)...$$

$$\coth z -\color{red}{\frac{1}{z}} = \color{red}{\dfrac{1}{z}} + \sum_{n\geq 1}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right) = \sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}\tag{5}$$

$$=\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}=\dfrac{2z}{\pi^2 n^2}\cdot \color{green}{\dfrac{1}{1+\left(\dfrac{z^2}{(\pi n)^2}\right)}}$$

We notice that the green part is a geometric series, with its formula $\dfrac{1}{1+t}=\sum{n=1}^{\infty}(-1)^{m-t}t^{m-t}$

$$\sum_{n=1}^{\infty}\dfrac{2z}{\pi^2 n^2}\cdot\sum_{m=1}^{\infty}(-1)^{m-t}\cdot \dfrac{z^{2m-2}}{(\pi n)^{2m-1}}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{2z^{2m-1}}{\pi^{2m}n^{2m}}=\sum_{m=1}^{\infty}\dfrac{2z^{2m-1}}{\pi^{2m}}\cdot\sum_{n=1}^{\infty}\dfrac{1}{n^{2m}}$$

And we know that $\sum_{n=1}^{\infty}\dfrac{1}{n^{2m}}=\zeta(2n)$

Thus we have:

$$\sum_{m=1}^{\infty}\dfrac{2z^{2m-1}\zeta(2n)}{\pi^{2m}}$$

Comparing the coefficients of $z^{2n-1}$

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

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

An interesting function that approximates values of Zeta function for both odd and even intergers

The following function, found in this thread of Mathstack, approximates many values of Zeta function at both odd and even intergers

$$f(s) = \frac{\pi^s}{\left\lfloor((2^s - 1)\frac{\pi^s}{2^s}\right\rfloor-1} \approx\zeta(s),$$

First few values are here:

$$\begin{array}{|c|c|c|} s & f(s) & |\zeta(s) - f(s)| \\ \hline2 &  \frac{\pi^2}{6} & 0 \\ \hline3 &  \frac{\pi^3}{26} & 0.00950\ldots \\ \hline4 &  \frac{\pi^4}{90} & 0 \\ \hline5 &  \frac{\pi^5}{295} & 0.00042\ldots \\ \hline6 &  \frac{\pi^6}{945} & 0 \\ \hline7 &  \frac{\pi^7}{2995} & 0.00009\ldots \\ \hline8 & \frac{\pi^8}{9450} & 0 \\ \hline9 & \frac{\pi^9}{29749} & 0.000011\ldots \\ \hline\end{array}$$

According to user247327, the approximation of $\frac{\pi^9}{29749}$ is $1.0020202$, which is very far from the assigned value on the table.