# Rigorous proof of Basel problem

I've read Euler's proof for Basel problem and I was wondering what were the details it was missing.

Based on that proof, this is my try, do you think it is correct or is any detail missing?

Thanks a lot for the replays.

Proof:

By Wierstrass factorization theorem we know that

$$\dfrac{\sin z}{z}={\displaystyle \prod_{n=1}^{\infty}\left( 1-\dfrac{z^2}{n^2\pi^2}\right)}$$

Then, if we define

$$P_k(z)={\displaystyle \prod_{n=1}^{k}\left( 1-\dfrac{z^2}{n^2\pi^2}\right)}$$

we can observe that

$$P_1(z)=1-\dfrac{z^2}{\pi^2}$$

$$P_2(z)=1-\dfrac{1}{\pi^2}\left( 1+\dfrac{1}{4}\right)z^2+\dfrac{1}{4\pi^4}z^4=1-\dfrac{1}{\pi^2}\left( 1+\dfrac{1}{4}\right)z^2+o(z^3)$$

So, lets proof by induction that

$$P_k(z)=1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{i=1}^{k} \dfrac{1}{i^2}}\right)z^2+o(z^3)$$

The base case is already proven so lets assume it is true for $$k$$, then we have that

$$P_{k+1}(z)={\displaystyle \prod_{n=1}^{k+1}\left( 1-\dfrac{z^2}{n^2\pi^2}\right)}={\displaystyle \prod_{n=1}^{k}\left( 1-\dfrac{z^2}{n^2\pi^2}\right)} \left(1- \dfrac{z^2}{(k+1)^2\pi^2}\right)=P_k(z) \left(1- \dfrac{z^2}{(k+1)^2\pi^2}\right)= \left( 1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{i=1}^{k} \dfrac{1}{i^2}}\right)z^2+o(z^3) \right) \left(1- \dfrac{z^2}{(k+1)^2\pi^2}\right)\\=1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{i=1}^{k+1} \dfrac{1}{i^2}}\right)z^2+\dfrac{1}{(k+1)^2\pi^4}\left( {\displaystyle \sum_{i=1}^{k} \dfrac{1}{i^2}}\right)z^4+\left(1- \dfrac{z^2}{(k+1)^2\pi^2}\right) o(z^3)\\=1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{i=1}^{k+1} \dfrac{1}{i^2}}\right)z^2+o(z^3)$$

because

$${\displaystyle \lim_{z \to 0} \dfrac{1}{z^3} \left( \dfrac{1}{(k+1)^2\pi^4}\left( {\displaystyle \sum_{i=1}^{k} \dfrac{1}{i^2}}\right)z^4+\left(1- \dfrac{z^2}{(k+1)^2\pi^2}\right) o(z^3) \right)}= {\displaystyle \lim_{z \to 0} \dfrac{1}{(k+1)^2\pi^4}\left( {\displaystyle \sum_{i=1}^{k} \dfrac{1}{i^2}}\right)z^3}+{\displaystyle \lim_{z \to 0}\left(1- \dfrac{z^2}{(k+1)^2\pi^2}\right) \dfrac{o(z^3)}{z^3}}=0$$

Then, we have that

$$\dfrac{\sin(z)}{z}={\displaystyle \lim_{k \to \infty} P_k}={\displaystyle \lim_{k \to \infty} \left(1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{i=1}^{k} \dfrac{1}{i^2}}\right)z^2+o(z^3) \right)}=1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}}\right)z^2+o(z^3) \hspace{1mm} (1)$$

On the other hand, by the MacLaurin series expansion of $$\sin(z)$$ and by the fact that the product of two Laurent series is another Laurent series with coefficients as specify here (Multiplying Laurent series), we get that the Taylor series for our function is

$$\dfrac{\sin(z)}{z}={\displaystyle \sum_{n=0}^\infty (-1)^n \dfrac{1}{(2k+1)!} z^{2k}}$$

Then, we can express

$$\dfrac{\sin(z)}{z}=1-\dfrac{1}{6} z^2 +o(z^3)$$

and because of (1) and the uniqness of Taylor's polynomials, we conclude that for every $$z \in \mathbb{C}$$

$$1-\dfrac{1}{\pi^2}\left( {\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}}\right)z^2 = 1-\dfrac{1}{6} z^2$$

from were we conclude that

$${\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2}}= \dfrac{\pi^2}{6}$$

Note:

In (1), I'm not sure if it is correct that

$${\displaystyle \lim_{k \to \infty} o(z^3)}=o(z^3)$$

My reasoning is that given $$\varepsilon >0$$, we know that there exists some $$\delta >0$$ such that if $$0<|z|<\delta$$

$$\left | \dfrac{o(z^3)}{z^3} \right| < \varepsilon, \hspace{1mm}$$

So, if $$0<|z|<\delta$$ we have that

$$\left | \dfrac{1}{z^3} {\displaystyle \lim_{k \to \infty} o(z^3)}\right|=\left | {\displaystyle \lim_{k \to \infty} \dfrac{o(z^3)}{z^3}}\right| = {\displaystyle \lim_{k \to \infty} \left | \dfrac{o(z^3)}{z^3}\right| } <\varepsilon$$

and then, $${\displaystyle \lim_{k \to \infty} o(z^3)}=o(z^3)$$.

• I think the bulk of the real work here is done by the Weierstraß factorization theorem (which was not available to Euler). – TonyK Jun 27 '19 at 17:36
• Actually knowing anything about the coefficient on any $z^n$ ($n \ge 3$) term is irrelevant so long as we know it converges; we already know the infinite product does, so we are essentially done. – Brevan Ellefsen Jun 27 '19 at 20:04
• The discussion about the term $o(z^3)$ indeed needs more rigor. Because as you multiply more and more terms, the hidden constant might go increasing and even diverge. – Yves Daoust Jun 28 '19 at 6:08
• @YvesDaoust It is true, and I'm not able to proof that the limit of the remainders I get from every $Pk(z)$ is in fact an $o(z^3)$ without assuming that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$. Any ideas of how to do it? – Eparoh Jun 28 '19 at 10:01
• @Eparoh you could do a bit of rearranging in the infinite product to show it converges as I allude to above (you already know the infinite product converges, since $\sin(z)$ is finite for all $z$, which quickly gives you that all coefficients must be finite with a little thought) or you could some inequality estimates, either on the product itself, by weakening the terms to a known product/series, etc. Try a couple things and respond with how your attempts have gone. – Brevan Ellefsen Jun 29 '19 at 22:51