TITLE: The sum of a series, continued
QUESTION [7 upvotes]: In this question the OP asks whether the sum
$$
f(q, \alpha) = \sum _{k=1}^{\infty } \frac{q^k \left(q^k-1\right)^\alpha}{(q;q)_k}
$$
is ever zero. An experiment with Mathematica indicates, to any reasonable precision that
$$f(2, 1) = \sum _{k=1}^{\infty } \frac{2^k }{(2;2)_{k-1}} = 0,$$
but I, for one, can't actually prove it. Is it true?

REPLY [6 votes]: We can use the Euler identity
$\sum_{k=0}^\infty \frac{x^k}{q^{k(k-1)/2}(1/q;1/q)_k}=\prod_{j=0}^\infty(1+q^{-j}x)$ 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.