In this post, the following two continued fractions discussed by Nicco are given,

$$A(q)= \left(\frac{\vartheta_2(0,q)}{2\,q^{1/4}}\right)^2= \cfrac{1}{1-q+\cfrac{q(1\color{red}-q)^2}{1-q^3+\cfrac{q^2(1\color{red}-q^2)^2}{1-q^5+\cfrac{q^3(1\color{red}-q^3)^2}{1-q^7+\ddots}}}}\tag1$$

$$B(q)=\frac{1}{2\,q^{1/2}}\frac{\vartheta_2(0,q^2)}{\vartheta_3(0,q^2)} =\cfrac{1}{1-q+\cfrac{q(1\color{blue}+q)^2}{1-q^3+\cfrac{q^2(1\color{blue}+q^2)^2}{1-q^5+\cfrac{q^3(1\color{blue}+q^3)^2}{1-q^7+\ddots}}}}\tag2$$

where $|q|<1$. One can see their beautiful affinity. On a hunch, I decided to investigate the similar cfrac,


Q1: Is it true that,

$$C(q) \overset{\color{red}?}= \frac{1}{q}\sum_{n=1}^\infty \frac{q^n}{1-q^{2n-1}} =\sum_{n=0}^\infty\sigma_0(2n+1)\,q^n = 1+2q+2q^2+2q^3+3q^4+\dots\tag4$$

where $\sigma_0(2n+1)$ is the divisor function? (I've checked it to hundreds of coefficients and it seems to hold. More info at A099774.) Note that the equality below is known,

$$\sum_{n=1}^\infty\sigma_0(n)\,q^n = \sum_{n=1}^\infty \frac{q^n}{1-q^n}$$

Q2: Can $C(q)$ be also expressed in terms of the Jacobi theta functions $\vartheta_k(0,q)$?

  • $\begingroup$ @ccorn: Good to know. I have a feeling that $C(q)$ is expressible by theta functions. See, for example, eq.6 of the odd divisor function. $\endgroup$ Aug 26, 2015 at 4:32
  • $\begingroup$ Regarding $\frac{1}{q}\sum_{n=1}^\infty \frac{q^n}{1-q^{2n-1}} \overset{?}=\sum_{n=0}^\infty\sigma_0(2n+1)\,q^n$, just set $q=r^2$ in the RHS and multiply by $r$, then the series is the odd part of the generating function for $\sigma_0(n)$. Rewrite that one using the known equality, massage the fractions to get rid of $(-1)^n$ in the denominators, then you end up with a sum whose parts are zero for even $n$. Then you can substitute back $r^2=q$, and you get the LHS. $\endgroup$
    – ccorn
    Aug 26, 2015 at 9:00
  • $\begingroup$ @ccorn: Thanks. I've removed the second question mark in $(4)$. That just leaves the first one. $\endgroup$ Aug 26, 2015 at 11:48
  • $\begingroup$ W. r. t. thetanulls: I have checked the usual suspects from my notes and from DLMF (you can search there for 1-q^{2n-1} which returns quite useful results), no match. Also some (squares of) Jacobi elliptic functions have turned up among the candidates, still no match. $\endgroup$
    – ccorn
    Aug 26, 2015 at 13:14
  • $\begingroup$ @ccorn: Since using the odd divisor function, we have $$\sum_{n=0}^\infty \sigma_1^{(o)} q^n = \sum_{n=0}^\infty \frac{n\, q^n}{1+q^n} = \frac{\vartheta_2^4(0,q) +\vartheta_3^4(0,q)}{24}$$ I just have the oddest wishful feeling that $C(q)$, like $A(q)$ and $B(q)$, should be expressible as well (though maybe more complicated). But thanks for checking your sources. $\endgroup$ Aug 26, 2015 at 13:51

2 Answers 2


The answer to Q1 is yes. About Q2, I don't know yet.

As usual, I suppose that $q=\exp(2\pi\mathrm{i}\tau)$ for $\tau\in\mathbb{H}$ (complex upper half plane) and define $$q_n = \exp\frac{2\pi\mathrm{i}\tau}{n}$$ Thus we can consider Theta functions to be functions of $\tau$. I write $\vartheta_k(0\mid\tau)$ instead of $\vartheta_k(0,q_2)$. This is mostly to avoid multivalue issues with $\vartheta_2$.

Proposition: For $q,r\in\mathbb{C}$ with $|q|<1$ and $|r|<1$, we have $$\cfrac{1}{1-q-\cfrac{r\,(1-q)^2}{1-q^3-\cfrac{rq\,(1-q^2)^2} {1-q^5-\cfrac{rq^2(1-q^3)^2}{1-q^7-\cdots}}}} = \sum_{n=0}^\infty \frac{r^n}{1-q^{2n+1}}\tag{P}$$ Then the claim for your Q1 follows from setting $r=q$.

Furthermore, the already settled claim for $A(q)$ follows from setting $r=-q$ because there is the known series expansion $$\left(\frac{\vartheta_2(0\mid2\tau)}{2q_4}\right)^2 = \sum_{n=0}^\infty q^n \sum_{d\mid(2n+1)} (-1)^{(d-1)/2} = \sum_{n=0}^\infty \frac{(-q)^n}{1-q^{2n+1}}$$ where the latter Lambert-like series can be understood as a simple sieving machinery that generates an odd $d$ in every step and updates the counters associated with all odd multiples of $d$.

I suspect that there are simpler ways of proving proposition (P), but in the current series of posts regarding such continued fractions (1st, 2nd, 3rd, 4th), everything starts with the formula from Ramanujan's 2nd notebook, chapter 16, entry 11: $$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty} = \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{*}$$ The original has the sign of $b$ flipped, but I like to emphasize the symmetry.

To obtain the continued fraction in (P), we would need $ab=-r$ and $a/b=-1$, which is fulfilled by $\{a,b\}=\{\pm\sqrt{r}\}$. However, this implies that $a+b=0$, and in that case, Ramanujan's formula gets reduced to $0=0$. We need to work around that.

We have already dealt with $a+b=0$, but only for $ab=q$, so we need a new twist here. First rewrite $$(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty = \overbrace{(-a;q)_\infty\,(a;q)_\infty}^{(a^2;q^2)_\infty} \left(\frac{(-b;q)_\infty}{(a;q)_\infty} -\frac{(b;q)_\infty}{(-a;q)_\infty}\right)$$ Then assume $0<|a|<1$ and recall the $q$-binomial theorem $$\frac{(-b;q)_\infty}{(a;q)_\infty} = \sum_{n=0}^\infty \frac{\bigl(-\frac{b}{a};q\bigr)_{n}}{(q;q)_{n}}\,a^{n}$$ which implies $$\begin{align} \frac{(-b;q)_\infty}{(a;q)_\infty}-\frac{(b;q)_\infty}{(-a;q)_\infty} &= \sum_{n=0}^\infty\frac{\bigl(-\frac{b}{a};q\bigr)_{n}}{(q;q)_{n}} \,(a^{n}-(-a)^n) \\ &= 2\sum_{n=0}^\infty\frac{\bigl(-\frac{b}{a};q\bigr)_{2n+1}}{(q;q)_{2n+1}} \,a^{2n+1} \\ &= 2(a+b) \sum_{n=0}^\infty\frac{\bigl(-\frac{b}{a}q;q\bigr)_{2n}}{(q;q)_{2n+1}} \,a^{2n} \end{align}$$ Voilà, we have a product of $(a+b)$ with a continuous function. Let's plug that into Ramanujan's formula $(*)$ and cancel the common factor $(a+b)$: $$\small\cfrac{1}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)} {1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}} = \frac{2\,(a^2;q^2)_\infty \sum_{n=0}^\infty\frac{\left(-\frac{b}{a}q;q\right)_{2n}}{(q;q)_{2n+1}}\,a^{2n}} {(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty}$$ This is continuous across $-\frac{b}{a}=1$, so let us see what happens there. The denominator of the right-hand side reduces to $2\,(-a;q)_\infty\,(a;q)_\infty=2\,(a^2;q^2)_\infty$ and therefore gets cancelled with the $q$-Pochhammer symbol in the numerator. The series simplifies as $$\sum_{n=0}^\infty\frac{(q;q)_{2n}}{(q;q)_{2n+1}}\,a^{2n} = \sum_{n=0}^\infty\frac{a^{2n}}{1-q^{2n+1}}$$ Thus we obtain $$\cfrac{1}{1-q-\cfrac{a^2(1-q)^2}{1-q^3-\cfrac{a^2q\,(1-q^2)^2} {1-q^5-\cfrac{a^2q^2(1-q^3)^2}{1-q^7-\cdots}}}} = \sum_{n=0}^\infty\frac{a^{2n}}{1-q^{2n+1}}$$ and now we can replace $a^2$ with $r$ and are almost done.

We had to assume $0<|a|<1$ along the way, which covers all $0<|r|<1$, but the case $r=0$ needs to be checked separately. Now $r=0$ causes both sides of (P) to reduce to $1/(1-q)$ because the $r^0$ in the series is to be properly interpreted as $1$ by continuity. Q.e.d.


Too long for a comment

first let's assume the following identity to be true $$\frac{1}{q}\sum_{n=1}^\infty \frac{q^n}{1-q^{2n-1}} = \cfrac{1}{1-q-\cfrac{q(1-q)^2}{1-q^3-\cfrac{q^2(1-q^2)^2}{1-q^5-\cfrac{q^3(1-q^3)^2}{1-q^7-\ddots}}}}$$

Now multiplying both sides by $(1-q)$ and letting $q\rightarrow1$, yields $$\lim_{q\rightarrow 1} \frac{1}{q}\sum_{n=1}^\infty \frac{q^n(1-q)}{1-q^{2n-1}} = \lim_{q\rightarrow1} {\cfrac{1}{1-\cfrac{\cfrac{q(1-q)}{1-q^2}}{\cfrac{1-q^3}{1-q^2}-\cfrac{\cfrac{q^2(1-q^2)}{1-q^3}}{\cfrac{1-q^5}{1-q^3}-\cfrac{\cfrac {q^3(1-q^3)}{1-q^4}}{\cfrac{1-q^7}{1-q^4}-\ddots}}}}}$$

Which after equivalence transformation becomes

$$\sum_{n=1}^\infty \frac{1}{2n-1} = \cfrac{1}{1-\cfrac{1^2}{3-\cfrac{2^2}{5-\cfrac{3^2}{7-\cfrac{4^2}{9 - \ddots}}}}}$$ Which is not a valid identity,since the arctan identity

$$\sum_{n=1}^\infty \frac{(-1)^{n-1}z^{2n-1}}{2n-1}=\cfrac{z}{1+\cfrac{(1z)^2}{3+\cfrac{(2z)^2}{5+\cfrac{(3z)^2}{7+\cfrac{(4z)^2}{9 + \ddots}}}}}$$

Is valid for $$|z|\leq1$$ ,$$z\ne i,-i$$

update :In fact this comment does not invalidate the conjectured identity,but rather strengthens it ,since both the series and cfrac approach the same singularity,as observed by Ccorn.Thus the cfrac does not have a finite limit but a singularity.

  • $\begingroup$ Nicco, don't create multiple accounts. It is forbidden. $\endgroup$ Sep 8, 2015 at 3:32
  • $\begingroup$ @ Tito Piezas :it was a mistake,thing is I'm not yet familiar with this forum,my apologies. $\endgroup$
    – Nicco
    Sep 8, 2015 at 3:47
  • $\begingroup$ @Tito piezas:what do you think of the limit? $\endgroup$
    – Nicco
    Sep 8, 2015 at 3:50
  • $\begingroup$ I think you may have used an assumption that leads to a conclusion different from mine and ccorn. Anyway, kindly delete this answer and new account, and just re-answer (if you still wish) using your correct account. $\endgroup$ Sep 8, 2015 at 4:13
  • $\begingroup$ Using $z=-\mathrm{i}q$, $q\to1$, both sides diverge, or rather, in $\bar{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$ both sides converge to $\infty$, so there is no contradiction. $\endgroup$
    – ccorn
    Sep 8, 2015 at 11:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .