How does this step
$$D(q)=\sum_{n=1}^\infty d(n)q^n$$
Become this step?
\begin{align} D(q) &=\sum_{n=1}^\infty\sum_{m|n}mq^n=\sum_{m=1}^\infty\sum_{m|n}mq^n \\ &=\sum_{m=1}^\infty\sum_{m|n}m(q^m+q^{2m}+q^{3m}+\cdots) \end{align}
2026-03-25 13:55:45.1774446945
Confused as to how this step in a number theory proof is performed
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The first $=$ writes $d(n)$ as $\sum_{m|n}m$. The second $=$ reverses the order of summation. The final expression is a misprint, as we shouldn't keep the $\sum_{m|n}$ once we expand the sum over $q^n$ terms as the expression in brackets.