# (z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)

## Theorem

The following formula holds: $$\dfrac{z}{1-q} {}_2\phi_1(q,q;q^2;z) = \displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{1-q^k},$$ where ${}_2\phi_1$ denotes basic hypergeometric phi.