I'm working through S.J. Patterson's "An introduction to the theory of the Riemann Zeta-Function" for fun, and trying to solve the exercises. However, I am already experiencing some trouble with exercise 1.3, specifically in trying to prove the last identity $$ \frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}=\sum_{m\ge 1}\sigma_a(m)\sigma_b(m)m^{-s} $$ under the assumptions $\Re(s) > 1$, $\Re(s-a)>1$, $\Re(s-b)>1$ and $\Re(s-a-b)>1$
My first instinct was to use the product representations of the zeta functions above the divisor line to simplify those, yielding (where $\mathbb{P}$ is the set of all primes) \begin{align} \zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b) &= \prod_{p\in\mathbb{P}} \left(1-p^{-s}\right)^{-1}\left(1-p^{-s}p^a\right)^{-1}\left(1-p^{-s}p^b\right)^{-1}\left(1-p^{-s}p^ap^b\right)^{-1}\\ &= \prod_{p\in\mathbb{P}} \left(\sum_{k\ge0} p^{-ks}\right)\left(\sum_{k\ge0} p^{-ks}p^{ka}\right)\left(\sum_{k\ge0} p^{-ks}p^{kb}\right)\left(\sum_{k\ge0} p^{-ks}p^{k\left(a+b\right)}\right)\\ &= \prod_{p\in\mathbb{P}} \sum_{k\ge0} \left(\sum_{l_1=0}^k p^{al_1}\right)\left(\sum_{l_2=0}^k p^{bl_2}\right)p^{-ks}\\ &= \prod_{p\in\mathbb{P}} \sum_{k\ge0} \sigma_a(p^k)\sigma_b(p^k)p^{-ks} \end{align}
However this last line seems to suggest the identity already holds without the divisor. What is the error I am making in the above reasoning?
Really what goes wrong here is a misconception of how products of sums work. The product (take $z = p^{-s}$ for convenience) $$ \left(\sum_{k\ge0} z^k \right)\left(\sum_{k\ge0} z^k p^{ka}\right)\left(\sum_{k\ge0} z^k p^{kb}\right)\left(\sum_{k\ge0} z^k p^{k\left(a+b\right)}\right) $$ is, as a power series in $z$, $$ \sum_{n \geq 0} c_n z^n $$ where $c_n$ is the Cauchy product/convolution $$ c_n = \sum_{i + j + k + l = n} 1^i p^{ja} p^{kb} p^{l(a+b)}. $$ This is distinctly different (for $n \geq 2$) to the attempted coefficients $$ \left(\sum_{l_1=0}^n p^{al_1}\right)\left(\sum_{l_2=0}^n p^{bl_2}\right) $$ since the latter will never have more than one copy of each power of $p$, whereas the former will: take e.g., for $n = 2$, $1 + 0 + 0 + 1 = 2$ and $0 + 1 + 1 + 0 = 2$.