(z/(1-q))2Phi1(q,q;q^2;z)=Sum z^k/(1-q^k)
From specialfunctionswiki
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.
Proof
References
- 1953: Arthur Erdélyi, Wilhelm Magnus, Fritz Oberhettinger and Francesco G. Tricomi: Higher Transcendental Functions Volume I ... (previous) ... (next): $4.8 (6)$ (typo in text: text has sum beginning at $k=0$)