I am studying the quotient of
 $$f(\varepsilon) = \sum\_{i=1}^{\infty} \frac{i^2}{2^{\varepsilon i^2}}$$
and $$g(\varepsilon) = \sum\_{i=1}^{\infty} \frac{1}{2^{\varepsilon i^2}}$$
for some $\varepsilon>0.$


Consider then the function $h(\varepsilon):=f(\varepsilon)/g(\varepsilon).$ 


Doing some numerics, I found that the second derivative of the function $h$ satisfies $\liminf\_{\varepsilon \downarrow 0} h''(\varepsilon)>0.$


I was not really able to quantify what the limiting expression is, but I am sure the second derivative stays positive as $\varepsilon$ tends to zero. 


Unfortunately, one cannot really evaluate the expressions, but perhaps there is some nice convexity trick to show that 


$$\liminf\_{\varepsilon \downarrow 0} h''(\varepsilon)>0$$


cause I do not really see where it comes from. 



Quotient with positive second derivative in the limit?
I expect for small $\epsilon$ the approximation of the sum by an integral to be accurate. Here is a plot of $h(\epsilon)$ (blue curve) and 
$$H(\epsilon)=\frac{\int\_0^\infty x^2 2^{-\epsilon x^2}\,dx}{\int\_0^\infty 2^{-\epsilon x^2}\,dx}=\frac{1}{\epsilon\log 4},$$
(gold curve).


![](https://ilorentz.org/beenakker/MO/ratioofsums.png)


This would imply that the second derivative diverges as $\epsilon^{-3}/\log 2$ for $\epsilon \rightarrow 0$.


