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I have computed a generating function for a problem involving a particular series, and would like to know if anyone has any references or a categorisation for it? It's $$ G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}. $$ It appears to be related to (mock) theta functions, but seems to be simpler. In particular, I would like to know whether $G(a,z)$ satisfies any identities?

Many thanks.

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    $\begingroup$ For searches: when you take "half" of a theta function like this, it may be called a "partial theta function". $\endgroup$ May 13, 2011 at 15:44

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It should be noted that using the Jacobi triple product that we have $$ H(a,z) = a^{-1}\sum_{n=-\infty}^\infty a^nz^{n(n+1)/2} = a^{-1}\prod_{m=1}^\infty (1 - z^m)(1 - z^ma)(1 + z^{m-1}a) $$ where the main difference is that the indexing shifts and we are doing about twice the sum that you are.

If we try relate these, we get $$ H(a,z) = \big(a^{-1} + G(a,z)\big) + a^{-2}\big(1 + a^{-1}G(a^{-1},z)\big). $$

I'm not sure how much this helps though.

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  • $\begingroup$ Thanks Simon, this is pretty much what I was hoping for! Although it doesn't seem possible to eliminate $G(a^{-1},z)$? $\endgroup$
    – Granger
    May 13, 2011 at 15:09
  • $\begingroup$ I'm not sure if it is. As this is really only 'half' of a 'theta function', I don't see an obvious way to eliminate that part, other than setting $a = 1$... $\endgroup$
    – Simon Rose
    May 13, 2011 at 17:50
  • $\begingroup$ The product is not quite right, did you have in mind$$\prod_{m=1}^\infty(1-z^m)(1+z^ma)(1+z^{m-1}/a)$$? $\endgroup$ Jul 25, 2015 at 6:29
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Your generating function is related to a simple continued fraction expansion due to Touchard:

$\sum\limits_{k \ge 0} ( - 1)^k q^{k+1\choose2} v^k $ =$ \frac{1}{{1 + v - \frac{{(1 - q)v}}{{1 + v - \frac{{(1 - q^2 )v}}{ \cdots }}}}}.$

A simple proof can be found in a paper by H. Prodinger http://de.arxiv.org/abs/1102.5186

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  • $\begingroup$ Thanks Johann, I hadn't seen this before, but I'm not sure whether it's helpful right now. $\endgroup$
    – Granger
    May 13, 2011 at 15:21
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Another possible connection is the following result of Gauss:

$$ \sum_{n=0}^\infty \ q^{n(n+1)/2} = \prod_{m=1}^\infty \ \frac{1-q^{2m}}{1-q^{2m-1}} $$

(Andrews The Theory of Partitions Corollary 2.10), actually a corollary of the Jacobi Triple Identity that Simon used.

One close mock theta function is $$\psi_0(q) = \sum_{n=0}^\infty \ q^{(n+1)(n+2)/2}(-q)_n$$

(see Andrews chapter 2 examples 12 and 13).

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