Closed form or upper bound for $\lim_{n\to\infty}\sum_{k=0}^n\frac{e^{-k^2x}-e^{-(k+1)^2x}}{2k+1} $?

Question

Consider the expression $$ f(x,n)=\sum_{k=0}^n\frac{e^{-k^2x}-e^{-(k+1)^2x}}{2k+1} $$ What is $\lim_{n\to\infty}f(x,n)$? If not a closed formula, is it possible to find an upper bound?

Note: This seems like some type of telescoping series, weighted by terms of the form $(2k+1)^{-1}$. The main reason I ask is that, if plotting $f$ for increasing values of $n$, we get what seems like an approximation to a limiting function of $x$, as seen here

Any ideas?

Answer 1

$$f(x)=\sum_{n=0}^\infty\frac{e^{-n^2x}-e^{-(n+1)^2x}}{2n+1}=\sum_{n\in\mathbb{Z}}\frac{e^{-n^2 x}}{2n+1}=\sum_{n\in\mathbb{Z}}\frac{e^{-n^2 x}}{1-4n^2}$$ is clearly related to the family of theta functions; we may express it in terms of $$\vartheta(x)=\sum_{n\in\mathbb{Z}}e^{-\pi(nx)^2},$$ chosen for its well-known property $\vartheta(1/x)=x\vartheta(x)$.

The last sum for $f(x)$ above (holds for $x\geqslant 0$ and) gives $$f(x)+4f'(x)=\sum_{n\in\mathbb{Z}}e^{-n^2 x}=\vartheta(\textstyle\sqrt{x/\pi})\\\implies f(x)=e^{-x/4}\int_0^{x/4}e^y\,\vartheta(2\textstyle\sqrt{y/\pi})\,dy.$$

In particular, for small $x$ we have $f(x)=\sqrt\pi D_+(\sqrt{x}/2)+O(xe^{-\pi^2/x})$, where $$D_+(z)=e^{-z^2}\int_0^z e^{t^2}\,dt=\frac12\sum_{n=0}^\infty\frac{(-1)^n n!}{(2n+1)!}(2z)^{2n+1}.$$ A less accurate estimate is then $f(x)=\sqrt{\pi x}/2+O(x^{3/2})$.

Answer 2

I cannot justify this rigourosly but let me try it for $x\geq 1+\varepsilon>1$ we have :

$$f\left(x\right)=\frac{e^{-k^{2}x}-e^{-(k+1)^{2}x}}{x\left(2k+1\right)}\leq \int_{0}^{\infty}\left(2k+1\right)f\left(x+1\right)dx$$

We use Frullani's integral to give a upper bound wich is :

$$2x\ln(n+1)$$

Following my idea we can improving it (and following the result @Metamorphy) if we have for $x\geq 1+\varepsilon>1$

$$g\left(x\right)=\frac{e^{-k^{2}x}-e^{-(k+1)^{2}x}}{\sqrt{x}\left(2k+1\right)}\leq C\int_{0}^{\infty}\frac{ke^{-k^{2}\left(1+x\right)}-\left(k+1\right)e^{-(k+1)^{2}\left(1+x\right)}}{\sqrt{1+x}}dx$$

We have as upper bound :

$$C\sqrt{x\pi}(\operatorname{erf(n)}-\operatorname{erf(1)})$$

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Now if we have for $x,k\geq 1$:

$$h\left(x\right)=C\int_{0}^{\infty}\left(k^{2}e^{-k^{2}\left(1+y\right)}-\left(k+1\right)^{2}e^{-(k+1)^{2}\left(1+y\right)}\right)dy-r\left(x\right)>0$$

Where $r\left(x\right)=\frac{e^{-k^{2}x}-e^{-(k+1)^{2}x}}{2k+1}$

We have as upper bound :

$$1$$

Starting from $k=2$ we have as upper bound :

$$\sum_{k=0}^{1}\frac{\left(e^{-k^{2}x}-e^{-\left(k+1\right)^{2}x}\right)}{2k+1}+e^{-4x}$$