THE series for $f(q,\alpha)$ must be alternating, otherwise the sum is strictly positive. Now, we can use the Euler Identity: $\sum_{k=0}^\infty \frac{x^k}{q^{k(k-1)/2}(1/q;1/q)_k}=\prod_{j=0}^\infty(1+q^{-j}x), q>1.$ If we take $x=-2$ and $q=2$, we obtain the series from the previous question and from the infinite product it will be 0.

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.

THE series for $f(q,\alpha)$ must be alternating, otherwise the sum is strictly positive. Now, we can use the Euler Identity: $\sum_{k=0}^\infty \frac{x^k}{q^{k(k-1)/2}(1/q;1/q)_k}=\prod_{j=0}^\infty(1+q^{-j}x), q>1.$ If we take $x=-2,$ we obtain the series from the previous question and from the infinite product it will be 0.

THE series for $f(q,\alpha)$ must be alternating, otherwise the sum is strictly positive. Now, we can use the Euler Identity: $\sum_{k=0}^\infty \frac{x^k}{q^{k(k-1)/2}(1/q;1/q)_k}=\prod_{j=0}^\infty(1+q^{-j}x), q>1.$ If we take $x=-2$ and $q=2$, we obtain the series from the previous question and from the infinite product it will be 0.