Currently I'm working through Murty's "Problems in the Theory of Modular Forms". He presents Theorem 1.3.1 as follows $$ \left(\sum_{-\infty}^{\infty}q^{n^2}\right)^2 = 1 + 4\sum_{n = 0}^{\infty} \left( \frac{q^{4n + 1}}{1 - q^{4n + 1}} - \frac{q^{4n + 3}}{1 - q^{4n + 3}}\right) $$ Now I've worked through the Theorem and I understand that $r_2(n)$, the number of ways that $n$ can be written as the sum of two squares, is given by the coefficient of $q^{n}$ on the right. However he then presents Corollary 1.3.2:
For $n \geq 1$, we have $r_2(n) = 4(d_1(n) - d_3(n))$ where $d_i(n)$ is the number of divisors of $n$ congruent to $i$ modulo $4$.
He then says that it follows immediately from expanding the summation on the right as a power series in $q$. Now I'm currently struggling to even write the summation on the right as a power series. I don't see how to eliminate the $q$s in the denominator. Any truncation of the series I compute via Mathematica still has $q$s in the denominator. Any insight or help would be amazing, thanks!.
$$1 + 4\sum_{n = 0}^{\infty} \left( \frac{q^{4n + 1}}{1 - q^{4n + 1}} - \frac{q^{4n + 3}}{1 - q^{4n + 3}}\right)=1+4\sum_{k=0}^\infty (\sum_{m=1}^\infty q^{(4k+1)m}-q^{(4k+3)m}) $$ $$ =1+4 \sum_{n=1}^\infty q^n (\sum_{4k+1 | n}1 -\sum_{4k+3 | n}1 )) =1+4 \sum_{n=1}^\infty q^n (d_1(n)-d_3(n))$$ Thus with $r_2(n) = \# \{ (a,b) \in \mathbb{Z}^2, a^2+b^2 = n\}$
$$\sum_{n=-\infty}^\infty r_2(n) q^n = \left(\sum_{n=-\infty}^{\infty}q^{n^2}\right)^2 = 1 + 4\sum_{n = 0}^{\infty} \left( \frac{q^{4n + 1}}{1 - q^{4n + 1}} - \frac{q^{4n + 3}}{1 - q^{4n + 3}}\right)$$ means $r_2(0) = 1$, $r_2(n) =4( d_1(n)-d_3(n))$