19:37
It seems none is interested in the question Mike asked (good, field's clear for me), so I'll write my epic solution here.
Suppose $f : \Bbb C \to \Bbb C$ is an entire function with simple zeroes $\lambda_i \neq 0$ and order of vanishing $n$ at $z = 0$. Take some other entire function $g$ with precisely those simple zeroes and order of vanishing at $z = 0$. $f/g$ is a nonzero entire function, so $f/g = e^h$ for some entire function $h$ by holomorphic lifting along the covering map $\exp : \Bbb C \to \Bbb C \setminus \{0\}$.
If $f$ has simple zeroes at $\lambda_i \neq 0$, then one usually constructs, as in the proof of Weierstrass factorization theorem, $$g(z) = z^n \prod_i E_{a_i}(z/\lambda_i)\;\text{where}\; E_{a_i}(z):=(1 - z) \exp (z + z^2/2 + \cdots + z^{a_i}/a_i)$$
We need to pick $\{a_i\}$ appropriately so that the product converges to an entire function. Notice that $E_{a_i}(z) := \exp(\log(1 - z) + z + z^2 +\cdots + z^{a_i}/a_i) = \exp(-\sum_{n > a_i} z^n/n)$. If $|z|< 1/2$, then the term inside the exponential is bounded by $2|z|^{a_i+1}$. Therefore $|1 - E_{a_i}(z)| = |1 - \exp(-\sum_{n > a_i} z^n/n)| \leq c|\sum_{n > a_i} z^n/n|<C|z|^{a_i + 1}$.
In particular, $|1 - E(z/\lambda_i)| \leq C |z|^{a_i+1}/|\lambda_i|^{a_i+1} = C |z/\lambda_i|^{a_i+1}$. If we can choose $a_i$ so that $\sum |\lambda_i|^{-1-a_i}$ converges we're done because by the simple lemma that $\prod (1 + a_n)$ converges whenever $\sum |a_n|$ converges.
Hadamard tightened what $g$ can be: Let's call the order of growth of $f$ to be $\rho = \inf \{r : |f(z)| \leq A\exp(B|z|^r) \; \forall z\in \Bbb C\}$. Then one can pick $h$ to be a polynomial of degree $\leq \lfloor \rho \rfloor$, according to the Hadamard factorization theorem.
@MikeMiller In our case, let's say $f(z) = \sin(z) - z\cos(z)$. This has order of growth $\rho = 1$. Notice that since $f$ is odd, the real zeroes of $f$ are $0, \pm \lambda_1, \pm \lambda_2, \cdots$.
I claim there are no non-real complex zeroes of $f$; this is just Rouche; consider the square $K = [-n\pi, n\pi]^2 \subset \Bbb C^2$. $|-z\cos(z)| < |\sin(z)|$ on $\partial K$, so $f$ has as many zeroes as $\sin$ in $\text{int} K$, which are exactly the integer multiples of $\pi$, and since there's a solution of $f(x) = 0$ between any two solutions of $\sin(x) = 0$ (except at $x = 0$, where those sets coincide), we're done by simply counting.
Since it's already given to me that $\sum \lambda_i^{-2}$ is finite (:p also easy to just prove by hand), I can take the constant sequence $a_i = 1$. The refined Weiestrass product theorem gives $$\sin(z) - z\cos(z) = e^{az+b} z^3 \prod_{k = 1}^\infty (1 + z/\lambda_k) e^{-z/\lambda_k}\prod_{k = 1}^\infty (1 - z/\lambda_k) e^{z/\lambda_k} = e^{az+b} z^3 \prod_{k = 1}^\infty (1 - z^2/\lambda_k^2)$$
The coefficient of $z^3$ in the Taylor expansion of $\sin(z) - z\cos(z)$ is $1/3$, and the coefficient of $z^4$ is $0$. Since the constant coefficient of the product term, when expanded out, is $1$, this forces $b = -\log(3)$ and $a = 0$.
Therefore $\displaystyle \sin(z) - z\cos(z) = z^3/3 \prod_{k = 1}^\infty (1 - z^2/\lambda_k^2)$
Compare coefficient of $z^5$ on both hands of this equality. One is, by writing down the Taylor expansion, $1/30$. The other is, by expanding out the product, $1/3 \sum_{k = 1}^\infty 1/\lambda_k^2$
$$\sum_{k = 1}^\infty \frac1{\lambda_k^2} = \frac3{30} = \frac1{10}$$
(This is of course an emulation of Euler's proof of the Basel problem)
