In this question the OP asks whether the sum $$ f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k1\right)^\alpha}{(q;q)_k} $$ is ever zero. An experiment with Mathematica indicates, to any reasonable precision that $$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k1}} = 0,$$ but I, for one, can't actually prove it. Is it true?
$$1/(21)\dots (2^{k1}1)+1/(21)\dots (2^k1)=2^k/(21)\dots (2^k1),$$ thus alternating sum of expression on the right is telescopical.
We can use the Euler identity $\sum_{k=0}^\infty \frac{x^k}{q^{k(k1)/2}(1/q;1/q)_k}=\prod_{j=0}^\infty(1+q^{j}x)$ for $q>1$. Taking $x=1$, we obtain the series $f(q,1)$ in the original post, and by the infinite product representation it equals 0.

$\begingroup$ Can you give a precise reference for this Euler identity? $\endgroup$ – GH from MO Sep 4 '15 at 18:53

1$\begingroup$ @GHfromMO, A quick way to prove it is to notice that the coefficient of $x^kq^{n}$ on both sides is the number of partitions of $n$ into $k$ distinct parts. $\endgroup$ – Gjergji Zaimi Sep 4 '15 at 18:58

1$\begingroup$ @GHfromMO One of the possible references is: G. E. Andrews, R. Askey, R. Roy, Special Functions, Cambridge Univ. Press, Cambridge 1999, Corollary 10.2.2 (b) on page 490. It is also presented in the Books by Kac&Cheung, and Gasper&Rahman, but I don't have the exact pages at the moment (Can provşde later on) $\endgroup$ – Deepti Sep 4 '15 at 19:10


$\begingroup$ I think we need to plug in $x=1$ instead of $x=2$, and this works for any $q>1$. This is because your sum on the left hand side equals $\sum_{k=0}^\infty\frac{(qx)^k}{(q;q)_k}$. $\endgroup$ – GH from MO Sep 4 '15 at 21:05