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.

Lipschitz's formula, which follows from Euler's partial fraction expansion of \pi \cot \pi z, gives G_2 = 2\zeta(2) + 2(2\pi i)^2 \sum_{n \geq 1} \sigma_1(n) q^n with \zeta(2) left unevaluated Definition 5.10; divide by 2\zeta(2).

This is the q-expansion of E_4 = G_4 / (2\zeta(4)), again with \zeta(4) left unevaluated Definition 5.10.

A power series can be differentiated term by term inside its disc of convergence, and D q^m = m q^m.

By Lemma 5.11, Lemma 5.12 and Lemma 5.13, the coefficients of q and q^2 are \alpha_2 and 6\alpha_2 in D E_2, 2\alpha_2 and 6\alpha_2 + \alpha_2^2 in E_2^2, and \alpha_4 and 9\alpha_4 in E_4, since \sigma_1(2) = 3 and \sigma_3(2) = 9. A convergent q-series determines its coefficients, so they agree on both sides.

\alpha_2 = -\frac{4\pi^2}{\zeta(2)} = -24 says \zeta(2) = \frac{\pi^2}{6} Definition 5.10, and \zeta(2) = \sum_{n \geq 1} \frac{1}{n^2} Definition 5.1.

Take Ramanujan's identity D E_2 = \frac{E_2^2 - E_4}{12} from mathlib. By Lemma 5.14 with \kappa = \frac{1}{12}, 12\alpha_2 = 2\alpha_2 - \alpha_4 and 72\alpha_2 = 6\alpha_2 + \alpha_2^2 - 9\alpha_4, so \alpha_4 = -10\alpha_2 and \alpha_2 (\alpha_2 + 24) = 0. As \alpha_2 \neq 0, \alpha_2 = -24 Lemma 5.15. This proof is not independent of mathlib's evaluation of \zeta(2): mathlib proves Ramanujan's identity using \zeta(2) = \frac{\pi^2}{6}, and the identity with the constant \frac{1}{12} is in fact equivalent to it, by the computation just made.

This is Euler's pentagonal number theorem \prod_{n \geq 1} (1 - q^n) = \sum_{k \in \mathbb{Z}} (-1)^k q^{k (3k - 1) / 2}, together with \frac{1}{24} + \frac{k (3k - 1)}{2} = \frac{(6k - 1)^2}{24}.

Write the series of Lemma 5.17 as e^{\pi i z / 12} \sum_k e^{-\pi a k^2 + 2\pi b k} with a = -3iz and b = \frac{i (1 - z)}{2}, and apply Poisson summation to this Gaussian: \sum_k e^{-\pi a k^2 + 2\pi b k} = a^{-1/2} \sum_n e^{-\pi (n + ib)^2 / a}. Expanding the square, the factor e^{\pi i z / 12} cancels.

Split n modulo 3. The classes n = 3j and n = 3j + 1 contribute e^{\pi i / 6} and e^{-\pi i / 6} times the series of Lemma 5.17 (the second after j \mapsto -j), and these factors add up to 2 \cos \frac{\pi}{6} = \sqrt{3}. The class n = 3j + 2 contributes a series that changes sign under j \mapsto -1 - j, so it vanishes.

By Lemma 5.18 and Lemma 5.19 at w = -1/z, \eta(z) = \sqrt{3}\,(-3iz)^{-1/2}\,\eta(-1/z). The 24th power removes the square root: \eta(z)^{24} = 3^{12} (-3iz)^{-12} \eta(-1/z)^{24} = z^{-12} \eta(-1/z)^{24}. Invariance under z \mapsto z + 1 and vanishing at i\infty are visible from the product.

\Delta vanishes nowhere on the upper half-plane and has a simple zero at the cusp, so dividing a cusp form of weight k by \Delta Lemma 5.20 gives a modular form of weight k - 12 < 0, which is zero.

For \gamma \in \mathrm{SL}_2(\mathbb{Z}) with bottom row (c, d), the transformation law of G_2 gives (cz + d)^{-2} E_2(\gamma z) = E_2(z) - \frac{2\pi i c}{2\zeta(2) (cz + d)}. Differentiating, the defects of D E_2 and of \kappa E_2^2 cancel exactly when \kappa \cdot \frac{(2\pi i)^2}{2\zeta(2)} = -1, that is, when \kappa = \frac{\zeta(2)}{2\pi^2}.

By Lemma 5.22, D E_2 - \kappa (E_2^2 - E_4) is modular of weight 4. It is holomorphic, and it tends to 0 - \kappa (1 - 1) = 0 at i\infty by Lemma 5.11 and Lemma 5.12. So it is a cusp form of weight 4, and it is zero by Lemma 5.21.

By Lemma 5.23 and Lemma 5.14 with \kappa = \frac{\zeta(2)}{2\pi^2}, \alpha_2 = \kappa (2\alpha_2 - \alpha_4) and 6\alpha_2 = \kappa (6\alpha_2 + \alpha_2^2 - 9\alpha_4), which are the note's equations with \frac{1}{12} replaced by \kappa. Since \kappa \alpha_2 = -2, multiplying both by \alpha_2 and eliminating \alpha_4 gives \alpha_2 (\alpha_2 + 24) = 0, so \alpha_2 = -24 Lemma 5.15.