# Showing that $\sum_{n=0}^\infty (4n+1)q^{\left (\frac{4n+1}{2}\right)^2} - \sum_{n=1}^\infty (4n-1)q^{\left (\frac{4n-1}{2}\right)^2} \geq 0.1$

Recently I came along the following problem concerning a lower bound on the difference of two series:

I want to show that for every $$q \in [e^{-2},e^{-\frac{1}{2}}]$$ we have

$$f(q) := \sum_{n=0}^\infty (4n+1)q^{\left (\frac{4n+1}{2}\right)^2} - \sum_{n=1}^\infty (4n-1)q^{\left (\frac{4n-1}{2}\right)^2} \geq \frac{1}{10}$$

If I plot the function $$f$$ on the interval $$[e^{-2},e^{-\frac{1}{2}}]$$ I get the following:

Hence, the minimum seems to be attained at $$q=$$ with $$f(e^{-\frac{1}{2}}) \approx 0.113$$. Does anyone has an idea or a simple approach how to show the above estimate?

• Letting $\chi_4$ be the unique non-principal Dirichlet character modulo $4$, your $f$ may be written as $\frac{1}{2} \times \sum_{n \in \mathbb{Z}} \chi_4(n) n q^{n^2/4} = \sum_{n \ge 1} \chi_4(n) n q^{n^2/4}$, which suggests that the theory of modular forms might be relevant. Perhaps $f$ factorizes? Sep 4, 2021 at 20:17
• @OfirGorodetsky It is $q^{1/4}\prod_{n\geqslant1}(1-q^{2n})^3$ Sep 5, 2021 at 6:01
• In other words, it is $\eta(q^2)^3$ Sep 5, 2021 at 6:07

Let $$s_q(N):=\sum_{n=0}^N f_q(n),\quad t_q(N):=\sum_{n=1}^N g_q(n),$$ where $$f_q(n):=(4n+1)q^{(4n+1)^2/4},\quad g_q(n):=(4n-1)q^{(4n-1)^2/4}.$$ We want to show that $$s_q(\infty)-t_q(\infty)\overset{\text{(?)}}\ge1/10 \tag{1}$$ for all $$q\in[e^{-2},e^{-1/2}]. \tag{2}$$

For such $$q$$, $$g_q(n)$$ is decreasing in $$n\ge1$$ and increasing in $$q$$, and hence $$t_q(\infty)-t_q(2) =\sum_{n=3}^\infty g_q(n) \\ <\int_2^\infty g_{e^{-1/2}}(u)\,du=e^{-49/8}.$$ So, for $$q$$ as in (2),
\begin{aligned} &s_q(\infty)-t_q(\infty) \\ &>s_q(2)-t_q(2)-e^{-49/8} \\ &=h(q):=9 q^{81/4}-7 q^{49/4}+5 q^{25/4}-3 q^{9/4}+q^{1/4}-e^{-49/8} \\ &\ge h(e^{-1/2})>1/10; \end{aligned} \tag{3} the penultimate inequality, $$h(q)\ge h(e^{-1/2})$$, in the above multiline display is easy to prove, since $$h(q)$$ is a simple polynomial in $$q^{1/4}$$ (see a proof below).

So, (1) indeed holds for all $$q$$ as in (2).

Proof of the inequality $$h(q)\ge h(e^{-1/2})$$ for $$q$$ as in (2): For $$u\in[e^{-1/2},e^{-1/8}]$$, let $$H(u):=h(u^4),\quad H_2(u):=\frac{H''(u)}{24u^7}, \quad H_3(u):=\frac{H_2'(u)}{80 u^{15}},$$ $$H_4(u):=\frac{H_3'(u)}{168 u^{23}}=729 u^{32}-49\le H_4(e^{-1/8})<0.$$ So, $$H_3$$ is decreasing (on $$[e^{-1/2},e^{-1/8}]$$), to $$H_3(e^{-1/8})>0$$. So, $$H_3>0$$ and hence $$H_2$$ is increasing, from $$H_2(e^{-1/2})<0$$ to $$H_2(e^{-1/8})>0$$.

So, for some $$c\in(e^{-1/2},e^{-1/8})$$, $$H$$ is concave on $$[e^{-1/2},c]$$ and convex on $$[c,e^{-1/8}]$$. Also, $$H(e^{-1/2})>H(e^{-1/8})$$ and $$H'(e^{-1/8})<0$$.

So, $$H(u)\ge H(e^{-1/8})$$ for $$u\in[e^{-1/2},e^{-1/8}]$$ -- that is, $$h(q)\ge h(e^{-1/2})$$ for $$q$$ as in (2). $$\quad\Box$$

• @J.Swail : Are you satisfied with this answer? Sep 10, 2021 at 0:21
• Yes, I like the answer very much. Thanks a lot!! Sep 10, 2021 at 18:05