# How may we find this limit?

$$\lim\limits_{x \to 0}\left[{\sum\limits_{n=0}^{\infty}\left(\frac{x}{e^{(nx)^2}}\right)}\right]$$ This is from an MMF thread.

http://www.mymathforum.com/viewtopic.php?f=22&t=45776

Thank you for your consideration in this matter.

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This is nothing more than the derivative of a Jacobi theta function. –  Lucian Jan 29 at 19:08
This is the theta itself, not any derivative. –  Balarka Sen Jan 29 at 19:59
Hi there Mr. TNT , as you can see i'm trying to understand this problem. :) –  neofoxmulder Jan 29 at 20:24
@neofoxmulder You really are a double agent, working both on MMF and MSE. =D –  Balarka Sen Jan 31 at 12:52

Let us only consider $x > 0$, the $x < 0$ case follows by parity.

The function $t\mapsto e^{-t^2}$ is strictly decreasing on $[0,\infty)$, hence

$$xe^{-(nx)^2} > \int_{nx}^{(n+1)x} e^{-t^2}\,dt.\tag{1}$$

Sum it up:

$$x\sum_{n=0}^\infty e^{-(nx)^2} > \int_0^\infty e^{-t^2}\,dt = \frac{\sqrt{\pi}}{2}.\tag{2}$$

On the other hand,

$$xe^{-((n+1)x)^2} < \int_{nx}^{(n+1)x} e^{-t^2}\,dt,$$

so

$$x\sum_{n=0}^\infty e^{-(nx)^2} = x + x\sum_{n=0}^\infty e^{-((n+1)x)^2} < x + \int_0^\infty e^{-t^2}\,dt.$$

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Well, seems my answer is unnecessary now. Note however the limit as $x\to 0^{-}$ is the opposite of the limit from the right. –  Pedro Tamaroff Jan 29 at 20:34
But you beat me by two minutes, ten seconds, leave it up. Yes, by parity we have two different one-sided limits. –  Daniel Fischer Jan 29 at 20:39
This is what I was trying to do in a very wrong way :) +1 –  Arkamis Jan 29 at 20:41
@ Arkamis , when you deleted your post all the comments got deleted too. @ Daniel Fischer , the limit DNE? @ anyone , is the M - test an If and only If? In other words , if the M - test fails then is uniform convergence of the series inconclusive by the M - test? Thank you for your replies and effort in this matter. –  neofoxmulder Jan 29 at 21:01
@neofoxmulder We have $\lim\limits_{x\to 0^+} x\sum e^{-(nx)^2} = \sqrt{\pi}/2$, and $\lim\limits_{x\to 0^-} x\sum e^{-(nx)^2} = -\sqrt{\pi}/2$ by parity, so the limit doesn't exist [if it existed, it could only be $0$]. The $M$-test is sufficient to conclude uniform (and absolute) convergence, but you can have uniform convergence also if the $M$-test fails, if the regions where each function is "large" play nicely together and don't overlap much. –  Daniel Fischer Jan 29 at 21:11
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Argue $$\lim_{\varepsilon \to 0^+}\sum_{n\geqslant 0}\varepsilon \exp(-\varepsilon^2n^2)=\int_0^{\infty}e^{-x^2}dx \tag 1$$

ADD I leave the following as a record after Daniel posted the proof if $(1)$, which anyone interested can mimic and prove

PROP Let $f: [0,\infty)\to [0,\infty)$ be a monotone decreasing integrable function over its domain. Then $$\lim_{\varepsilon \to 0^+}\sum_{n\geqslant 0}\varepsilon f(\varepsilon n)=\int_0^{\infty}f$$

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You are an evil genius ! :-) –  Lucian Jan 29 at 20:22