Prove that $r_4(n) = 8\sum_{d|n,4\nmid d} d$
using Ramanujan's1 $\psi$1 formula.
I am a little stuck here and as just wondering if I could get some advice. My work with Jacobis triple product is a little rusty. Should I find out how to do $r_2$ first and work off of it?
There is a nice and detailed proof of it in this paper of Cooper and Hirschhorn; see Lemma $2$ on page $14$ (which uses Lemma 1 on the same page). The authors prove as well formulas for $r_6(n), r_8(n), r_{10}(n), r_{12}(n)$, and for $r_3(n), r_5(n), r_{7}(n), r_{9}(n), r_{11}(n)$. In particular $$ r_4(n)= 8\sum_{d\mid n, 4\nmid d} d = 8\sum_{d\mid n} d -32\sum_{d\mid \frac{n}{4}}d. $$ For a proof of Ramanujan's general formula for $\phi(q)^{2k}$, see the artilce of Cooper, S.: On sums of an even number of squares, and an even number of triangular numbers: an elementary approach based on Ramanujan’s $1\psi_1$ summation formula.