New(?) proof of Bassel problem

In this post, we give a new short proof of the Bassel problem $\zeta(2) = \pi^2 / 6$ using Eisenstein series.

For even $k \geq 2$, the Eisenstein series of weight $k$ and level $1$ is defined as

\[\begin{align*} G_k(z) &= \frac{1}{2} \sum_{(m, n) \in \mathbb{Z}^2 \backslash \{(0, 0)\}} \frac{1}{(mz + n)^k} \\ & =\frac{1}{2} \sum_{n \neq 0} \frac{1}{n^k} + \frac{1}{2} \sum_{m \neq 0} \sum_{n \in \mathbb{Z}} \frac{1}{(mz + n)^k}. \end{align*}\]

They are $1$-periodic, and we can compute their Fourier coefficients using Lipschitz’s formula:

\[\sum_{n \in \mathbb{Z}} \frac{1}{(z + n)^k} = \frac{(2 \pi i )^k}{(k-1)!} \sum_{r =1}^{\infty} r^{k-1} q^{k}\]

for $q = e^{2 \pi i z}$, which can be derived from Euler’s identity

\[\sum_{n \in \mathbb{Z}} \frac{1}{z + n} = \frac{\pi}{\tan \pi z} \quad (z \in \mathbb{C} \backslash \mathbb{Z})\]

by differentiating it $(k-1)$-times. Using this, we can prove that $G_k(z)$ has the following Fourier expansion (or $q$-expansion)

\[G_k(z) = \zeta(k) + \frac{(2\pi i )^{k}}{(k-1)!} \sum_{n\geq 1} \sigma_{k-1}(n) q^n.\]

See Zagier’s famous expository article Elliptic modular forms and applications on 1-2-3 of modular forms, Proposition 5 on page 16. Note that the first series defining $G_k(z)$ absolutely converges only if $k > 2$, and we use the second series for $k = 2$, which still admits the same form of the Fourier expansion One can normalize these forms as

\[E_k(z) = 1 + \frac{(2 \pi i )^k}{\zeta(k)(k-1)!} \sum_{n\geq 1} \sigma_{k-1}(n) q^n = 1 + \alpha_{k} \sum_{n\geq 1} \sigma_{k-1}(n) q^n\]

for $\alpha_k = \frac{(2 \pi i)^k}{\zeta(k) (k-1)!}$. Then our goal is reduced to compute $\alpha_2$ (and maybe other $\alpha_k$’s, too).

The key idea is to use the Ramanujan’s identity:

\[E_2' = \frac{E_2^2 - E_4}{12}, \quad E_4' = \frac{E_2 E_4 - E_6}{3}, \quad E_6' = \frac{E_2 E_6 - E_4^2}{2}.\]

Here the derivatives are $F’ := q \frac{dF}{dq} = \frac{1}{2 \pi i} \frac{dF}{dz}$. Especially, we’ll use the first identity. Comparing their coefficients of $q$ and $q^2$ in $q$-expansions gives two equations

\[\begin{align*} \alpha_2 &= \frac{2 \alpha_2 - \alpha_4}{12} \\ 6 \alpha_2 &= \frac{6\alpha_2 + \alpha_2^2 - 9 \alpha_4}{12} \end{align*}\]

and solving these equations gives $\alpha_2 = -24$ and $\alpha_4 = 240$ (note that $\alpha_2$ and $\alpha_4$ are nonzero by definition). Unwinding the definition of these numbers give

\[\begin{align*} \alpha_2 = -24 = \frac{(2\pi i)^2}{\zeta(2) \cdot 1} &\Leftrightarrow \zeta(2) = \frac{\pi^2}{6}, \\ \alpha_4 = 240 = \frac{(2 \pi i)^4}{\zeta(4)\cdot 6} &\Leftrightarrow \zeta(4) = \frac{\pi^4}{90}. \end{align*}\]

Other identities on Eisenstein series may give values of other $\zeta(k)$ for even $k \geq 6$. For example, one can compute $\zeta(6)$ using the second or third Ramanujan’s identity. The value of $\zeta(8)$ can be obtained via $E_4^2 = E_8$.

Tags: math