View source for (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]].

==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Arthur Erdélyi|author2=Wilhelm Magnus|author3=Fritz Oberhettinger|author4=Francesco G. Tricomi|prev=1Phi0(a;;z)1Phi0(b;;az)=1Phi0(ab;;z)|next=2Phi1(q,-1;-q;z)=1+2Sum z^k/(1+q^k)}}: $4.8 (6)$ (typo in text: text has sum beginning at $k=0$)

[[Category:Theorem]]
[[Category:Unproven]]