For what infinite series can a closed form be obtained by means of the $\text{Sum} = \text{Product} $ method? Euler solved the Basel problem by equating the Taylor series and the infinite product representation of $\sin(x)/x$:
$$\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2n}}{(2n+1)!} = \prod_{k=1}^{\infty}\bigg{(} 1 - \frac{x^{2}}{k^{2}\pi^{2}} \bigg{)} .$$ He proceeded to multiply out the product on the right-hand side and collecting all the $x^{2}$ terms. This coefficient amounts to: $$-\frac{1}{\pi^{2}}\sum_{q=1}^{\infty}\frac{1}{q^{2}}. $$  At the same time, the $x^{2}$ coefficient on the left-hand side equals $-\frac{1}{3!} = - \frac{1}{6} $. Equating them gives $$- \frac{1}{6} = -\frac{1}{\pi^{2}}\sum_{q=1}^{\infty}\frac{1}{q^{2}},$$ which yields $$\zeta(2) = \sum_{q=1}^{\infty}\frac{1}{q^{2}} = \frac{\pi^{2}}{6} .$$
In a similar manner, other zeta values at even arguments can be obtained, which Euler did for $\zeta(2), \zeta(4), \dots, \zeta(26)$.
One might be tempted to try something similar to obtain closed forms for the zeta values at odd arguments as well. However, this doesn't work -  see for instance this and that MO question.
I'll call the method I just described - equating the Taylor series and infinite product representation of a function, extract likewise coefficients from both sides, and equating these two again - the 'Sum = Product method' to obtain closed forms of infinite series.
Besides the sinc function, there are many other functions which have both a Taylor series and an infinite product expression - see for instance this overview over at Mathworld for some of the latter cases.
Question: what infinite series - besides $\zeta(2k)$ for $k\geq 1$ - can be evaluated in closed form by means of the Sum = Product method?
 A: Define $$ f(x) = \prod_{k=1}^{\infty} \left( 1 - g(k) \: x \right) = \sum_{m=0}^{\infty} h(m) \: x^m $$
then $$ \ln(f(x)) = \sum_{k=1}^{\infty}  \ln \left( 1 - g(k) \: x \right) \\  \ln(f(x)) = -\sum_{k=1}^{\infty}  \sum_{n=1}^{\infty}  g^n(k) \frac{x^n}{n} \\
 \ln(f(x)) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \sum_{k=1}^{\infty}  g^n(k) \\
f(x) = \text{Exp}\left(-\sum_{n=1}^{\infty} \frac{x^n}{n} \sum_{k=1}^{\infty}  g^n(k) \right) $$
This may look useless, but utilizing recurrence relation $$ q(n) = \frac{1}{n} \sum_{k=0}^{n-1} r(n-k) \cdot q(k) \: \: ; \: \: q(0) = 1\\\sum_{n=0}^{\infty} q(n) x^n = \text{Exp}\left( \sum_{k=1}^{\infty} r(k) \frac{x^k}{k} \right)$$
shows that $$ h(m) = \frac{1}{m} \sum_{k=0}^{m-1} \left(\sum_{j=1}^{\infty}  g^{m-k}(j) \right) \cdot h(k) $$
For example:$$ -\pi ^2 (x-1) \sqrt{x} \csc \left(\pi  \sqrt[4]{x}\right) \text{csch}\left(\pi  \sqrt[4]{x}\right) = \prod_{k=1}^{\infty} \left( \frac{1}{1-\frac{x}{(k+1)^4}}\right) $$ therefore $$ q(n) = \frac{1}{n} \sum_{k=0}^{n-1} \zeta(4(n-k)) \cdot q(k)  $$
But wait, you posited without $\zeta(s)$. Well the trouble is that solving the sum below is difficult for a closed form. $$ \left(\sum_{j=1}^{\infty}  g^{m-k}(j) \right)  $$ except for the trivial case: (as used in the example above) $$ \left(\sum_{j=1}^{\infty}  \frac{1}{j^{m-k}} \right) $$ And unless one has a closed form for f(x) or the sum on the right hand side, you won't be able to equate any coefficients.
