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The Basel problem is a famous problem that was posed in 1644 and solved by the genius mathematician Leonard Euler in 1734. The problem requires the sum of the infinite series:
$$1 + \frac 1 {2^2} + \frac 1 {3^2} + \frac 1 {4^2} + \dots = ?$$
Euler later proved that this sum is equal to \(\displaystyle \frac {\pi^2} 6\).
If you search for the string “Basel problem”, you will find many different ways to prove this problem. While wandering around the internet, I accidentally found a quite interesting and “beautiful” elementary proof for this problem by R J Ransford in Eureka magazine No. 42, published in the summer of 1982. Below is the note of the solution.
De Moivere's formula
According to Euler: \(e^{ix} = \cos x + i\sin x\)
Then: \((e^{ix})^n = (\cos x + i \sin x)^n = e^{i(nx)} = \cos nx + i \sin nx\)
Implies: \(\boxed{\cos nx + i \sin nx = (\cos x + i \sin x)^n}\)
Expanding the right side of the formula we have:
$$\begin{align*} \cos nx + i\sin nx &= \binom{n}{0}\cos^nx + \binom {n}{1}cos^{n-1}.i.\sin x + \dots \\ &=\left(\binom{n}{0}\cos^nx + \binom{n}{2}\cos^{n-2}x.i^2.\sin^2x + \dots + \dots \right) \\ &+ i\left(\binom{n}{1}\cos^{n-1}x.\sin x + \binom{n}{3}\cos^{n-3}x. i^2.\sin^3 x + \binom{n}{5}\cos^{n-5}x. i^4.\sin^5 x + \dots\right)\\ &= \left(\binom{n}{0}\cos^nx - \binom{n}{2}\cos^{n-2}x.\sin^2x + \dots \right) \\ &+ i\left(\binom{n}{1}\cos^{n-1}x.\sin x - \binom{n}{3}\cos^{n-3}x.\sin^3 x + \binom{n}{5}\cos^{n-5}x.\sin^5 x - \dots\right) \end{align*}$$
Comparing the imaginary parts of the right and left sides, we deduce:
$$\boxed{\sin nx = \binom{n}{1}cos^{n-1}x\sin x - \binom{n}{3}cos^{n-3}x\sin^3 x + \binom{n}{5}cos^{n-5}x\sin^5 x - \dots}$$
Vieta's formula for higher degree polynomials
Given a polynomial of degree \(n\) with coefficients \(a_i \in \mathbb C\) \(P(x) = a_nx^n+a_{n-1}x^{n-1}+\dots+a_0\) and those complex roots \(r_n,r_{n-1},\dots,r_1\), then for all \(0\le k\le n\), we have:
$$\displaystyle \sum_{1 \le i_1 \le i_2 \le \dots \le n} r_{i_1}r_{i_2}\dots r_{i_k} = (-1)^k \frac {a_{n-k}} {a_n}$$
Specific case of the above formula:
$$\boxed{ \sum_{i=1}^n r_i = - \frac {a_{n-1}}{a_n}}$$
The proof
First, let's prove the formula:
$$\cot^2(\frac {\pi} {2m+1}) + \cot^2(\frac {2\pi}{2m+1}) + \dots + \cot^2(\frac {m\pi} {2m+1}) = \frac 1 3 m (2m-1), (\forall m \ge 1)$$
Indeed, reapplying the above \(\sin nx\) formula:
$$\sin nx = \binom{n}{1}cos^{n-1}x\sin x - \binom{n}{3}cos^{n-3}x\sin^3 x + \binom{n}{5}cos^{n-5}x\sin^5 x - \dots$$
Let \(n=2m+1\) and \(\displaystyle x=\frac {r\pi}{2m+1} , (r=1,2,\dots,m),\) when \(\sin nx = 0, \sin x \gt 0\), devide both sides to \(\sin ^nx\):
$$0 = \binom{2m+1}{1}\cot^{2m}x - \binom{2m+1}{3}\cot^{2m-2}x + \dots$$
Let \(\displaystyle t=\cot^2x\), the formula becomes a polynomial with parameter \(t\)
$$P(t)= \binom{2m+1}{1}t^m - \binom{2m+1}{3}t^{m-1} + \dots$$
\(P(t) = 0\) then \(\displaystyle t=\cot^2(\frac{r\pi}{2m+1}), (r=1,2,\dots,m)\). Applying Vieta's formula to the sum of the above solutions, we can calculate the value of the sum of the solutions as \(\displaystyle \frac {\binom{2m+1}3}{\binom{2m+1}1} = \frac 1 3 m(2m-1)\). We have proven the formula:
$$\cot^2(\frac {\pi} {2m+1}) + \cot^2(\frac {2\pi}{2m+1}) + \dots + \cot^2(\frac {m\pi} {2m+1}) = \frac 1 3 m (2m-1), (\forall m \ge 1)$$
Add \(m\) to both sides:
$$\begin{align*} m + \cot^2(\frac {\pi} {2m+1}) + \cot^2(\frac {2\pi}{2m+1}) + \dots + \cot^2(\frac {m\pi} {2m+1}) &= m+ \frac 1 3 m (2m-1) \\ (1 + \cot^2(\frac {\pi} {2m+1})) + (1+\cot^2(\frac {2\pi}{2m+1})) + \dots + (1+\cot^2(\frac {m\pi} {2m+1})) &= \frac 1 3 m(2m+2) \end{align*}$$
Apply \(\csc^2y = 1+ \cot^2y\):
$$\csc^2(\frac {\pi} {2m+1}) + \csc^2(\frac {2\pi}{2m+1}) + \dots + \csc^2(\frac {m\pi} {2m+1}) = \frac 1 3 m(2m+2)$$
Seeing that when \(0 \lt y \lt \pi/2\) then \(0 \lt \sin y \lt y \lt \tan y,\) therefore \(\csc y \gt 1/y \gt \cot y\):
$$\begin{align*} \displaystyle \frac 1 3m(2m-1) &\lt (\frac {2m+1} {\pi})^2 + (\frac {2m+1} {2\pi})^2 + \dots + (\frac {2m+1} {m\pi})^2 &\lt \frac 1 3 m(2m+2) \\ \iff \frac 1 3m(2m-1) &\lt \frac {(2m+1)^2} {\pi^2} + \frac {(2m+1)^2} {2^2\pi^2} + \dots + \frac {(2m+1)^2} {m^2\pi^2} &\lt \frac 1 3 m(2m+2) \end{align*}$$
Multiply all terms to \(\frac{\pi^2} {(2m+1)^2}\) to get:
$$\begin{align*} \displaystyle \frac {\pi^2} 3 \left[ \frac 1 2 (1-\frac 1 {2m+1}) (1-\frac 2 {2m+1}) \right] \lt 1+\frac 1 {2^2} + \frac 1 {3^2} + \dots + \frac 1 {m^2} \lt \frac {\pi^2} 3 \left[ \frac 1 2 (1-\frac 1 {2m+1}) (1+\frac 1 {2m+1})\right] \\ \iff \frac {\pi^2} 6 (1-\frac 1 {2m+1}) (1-\frac 2 {2m+1}) \lt 1+\frac 1 {2^2} + \frac 1 {3^2} + \dots + \frac 1 {m^2} \lt \frac {\pi^2} 6 (1-\frac 1 {2m+1}) (1+\frac 1 {2m+1}) \\ \iff \frac {\pi^2} 6 (1-\frac 2 {2m+1}) \lt 1+\frac 1 {2^2} + \frac 1 {3^2} + \dots + \frac 1 {m^2} \lt \frac {\pi^2} 6 (1+\frac 1 {2m+1}) \end{align*}$$
When \(m \to \infty\), then \(\displaystyle \frac 1 {2m+1} \to 0\), \(\displaystyle \frac 2 {2m+1} \to 0\). We conclude:
$$\boxed{1+\frac 1 {2^2} + \frac 1 {3^2} + \dots = \frac {\pi^2} 6}$$