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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

enter image description here

Any ideas?

Blue
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sam wolfe
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  • What is $k$? Is that meant to be $i$? – Joshua Tilley Jan 31 '23 at 18:03
  • @JoshuaTilley yes! just corrected it – sam wolfe Jan 31 '23 at 18:04
  • The series looking like a telescoping series is no guarantee that it should be easy to calculate. $\int (e^x / x) dx $ looks like $\int e^x$, but the former is not an elementary function, while the latter most certainly is. – Charles Hudgins Feb 10 '23 at 01:12
  • Note that the series definitely converges and converges absolutely. It is bounded above by $\sum_k \frac{1}{e^{k^2 x}(2k + 1)}$, which converges by, say, the ratio test for all $x$. – Charles Hudgins Feb 10 '23 at 01:17
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    Splitting the sums still makes each one converge and $\sum\limits_{k=0}^\infty (2k+1) e^{((2k+1)a-(k+1/2)^2x}$ is a derivative of a theta function, so $\sum\limits_{k=0}^\infty\frac{e^{-(k+a)^2x}}{2k+1}$ could be an integral of a theta function. – Тyma Gaidash Feb 10 '23 at 01:19
  • It seems like $\lim_{n \to \infty} f\left( x,, n \right) \approx -e^{-x} + 1$ and $\lim_{n \to \infty} f\left( x,, n \right) \approx -\lim_{n \to \infty} \frac{\operatorname{d}}{\operatorname{d}x} f\left( x,, n \right) + 1$ applies. I don't know if that's helpful, but I wasn't expecting that. – The Art Of Repetition Feb 10 '23 at 03:44

2 Answers2

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$$f(x)=\sum_{n=0}^\infty\frac{e^{-n^2x}-e^{-(n+1)^2x}}{2n+1}=\sum_{n\in\mathbb{Z}}\frac{e^{-n^2 x}}{2n+1}=\sum_{n\in\mathbb{Z}}\frac{e^{-n^2 x}}{1-4n^2}$$ 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 $f(x)=\sqrt{\pi x}/2+O(x^{3/2})$.

metamorphy
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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 :

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

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 :

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

Barackouda
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