Let f(x) = x on the interval (-\pi, \pi], viewed as a square-integrable function on the circle \mathbb{R} / 2\pi\mathbb{Z}. Integration by parts gives its Fourier coefficients c_n; for n \neq 0 the boundary term contributes c_n = \frac{(-1)^n}{2\pi i n} \cdot 2\pi, whose squared modulus is \lvert c_n \rvert^2 = 1/n^2, while c_0 = 0 because f is odd. Hence the identity \lvert c_0 \rvert^2 = 1/0^2 = 0 makes \lvert c_n\rvert^2 = 1/n^2 hold for every integer n. Parseval's identity states \sum_{n \in \mathbb{Z}} \lvert c_n \rvert^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} \lvert f(x) \rvert^2 \, dx. The right-hand side is \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \, dx = \frac{1}{2\pi} \cdot \frac{2\pi^3}{3} = \frac{\pi^2}{3}. The left-hand side is \sum_{n \in \mathbb{Z}} \frac{1}{n^2} = 2 \sum_{n \geq 1} \frac{1}{n^2} because the summand is even and the n = 0 term vanishes. Equating the two sides gives 2 \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{3}, hence \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6}, the value asserted by the Basel problem Definition 5.1.

Since \cos\theta + i\sin\theta = e^{i\theta}, we have \sin((2n+1)\theta) = \operatorname{Im}\big((\cos\theta + i\sin\theta)^{2n+1}\big). Expanding the power by the binomial theorem and taking the imaginary part keeps exactly the terms with an odd power of i\sin\theta; reindexing those terms by j gives the stated sum.

Divide the De Moivre expansion Lemma 5.3 by \sin^{2n+1}\theta. Each summand becomes (-1)^j \binom{2n+1}{2j+1} \cos^{2(n-j)}\theta / \sin^{2(n-j)}\theta = (-1)^j \binom{2n+1}{2j+1} (\cot^2\theta)^{n-j}, and the sum of these is exactly P_n(\cot^2\theta) Definition 5.4.

For k = 1, \dots, n the angle \theta_k = \frac{k\pi}{2n+1} lies in (0, \pi/2) and satisfies \sin((2n+1)\theta_k) = \sin(k\pi) = 0 while \sin\theta_k \neq 0, so Lemma 5.5 shows each \cot^2\theta_k is a root of P_n. The n values \cot^2\theta_k are distinct, hence are exactly the roots of the degree-n polynomial P_n. By Vieta's formula their sum is the negative ratio of the two leading coefficients, \binom{2n+1}{3} / (2n+1) = \frac{n(2n-1)}{3}.

From x < \tan x on (0, \pi/2) we get \cot x < \frac{1}{x}; both sides are positive, so squaring preserves the inequality.

From 0 < \sin x < x we get \frac{1}{x^2} < \frac{1}{\sin^2 x} = 1 + \cot^2 x.

Apply the squeeze \cot^2 x < \frac{1}{x^2} < 1 + \cot^2 x (Lemma 5.7 and Lemma 5.8) at the n angles \theta_k = \frac{k\pi}{2n+1} and sum over k. Using \sum_{k=1}^n \frac{1}{\theta_k^2} = \frac{(2n+1)^2}{\pi^2} \sum_{k=1}^n \frac{1}{k^2} together with Cauchy's identity Lemma 5.6 gives \frac{n(2n-1)}{3} < \frac{(2n+1)^2}{\pi^2} \sum_{k=1}^n \frac{1}{k^2} < \frac{n(2n-1)}{3} + n. Multiplying through by \frac{\pi^2}{(2n+1)^2} sandwiches the partial sum \sum_{k=1}^n \frac{1}{k^2} between two sequences both tending to \frac{\pi^2}{6}, so by the squeeze theorem the series converges to \frac{\pi^2}{6}, the value asserted by the Basel problem Definition 5.1.