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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?

Community
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Igor Rivin
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2 Answers2

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$$1/(2-1)\dots (2^{k-1}-1)+1/(2-1)\dots (2^k-1)=2^k/(2-1)\dots (2^k-1),$$ thus alternating sum of expression on the right is telescopical.

Fedor Petrov
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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.

GH from MO
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Deepti
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  • Can you give a precise reference for this Euler identity? – GH from MO Sep 04 '15 at 18:53
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    @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. – Gjergji Zaimi Sep 04 '15 at 18:58
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    @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) – Deepti Sep 04 '15 at 19:10
  • @Deepti: Thank you! – GH from MO Sep 04 '15 at 19:29
  • 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 $\sum_{k=0}^\infty\frac{(-qx)^k}{(q;q)_k}$. – GH from MO Sep 04 '15 at 21:05