Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

Thank you for your consideration in this matter.

share|improve this question
    
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
add comment

2 Answers

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

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

share|improve this answer
    
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
1  
@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
show 1 more comment

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 $$\lim_{\varepsilon \to 0^+}\sum_{n\geqslant 0}\varepsilon f(\varepsilon n)=\int_0^{\infty}f $$

share|improve this answer
    
You are an evil genius ! :-) –  Lucian Jan 29 at 20:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.