$q$ 級数のまとめ

$$ \begin{aligned}\sum_{n=0}^\infty \frac{q^{n^2}}{(q; q)n}&=\frac{1}{(q,q^4; q^5)\infty}, \\[1em]\sum_{n=0}^\infty \frac{q^{n(n+1)}}{(q; q)n}&=\frac{1}{(q^2,q^3; q^5)\infty}.\end{aligned} $$

q-Series