2

I'm aware that this sequence converges to 0, but I'd like to know why my first instinct was wrong. I looked at this and took a cue from the old 1^inf technique of raising e to the natural log of the sequence:

$$\lim_{n\to \infty} n^{-n^2} \implies e^{\ln(n^{-n^2})}$$

leave e aside for now and concentrate on the natural log problem

$$-n^2\cdot \ln n\implies \frac{\ln n}{-n^{-2}}$$

by l'Hôpital's rule

$$\frac{\frac1n}{2n^{-3}} =\frac{n^2}2 $$

back to the original problem

$$\lim_{n\to \infty} e^{n^2/2} =( e^\infty)= \infty$$

The book offers a squeeze/sandwich theorem explanation, but I need to know what my mistake is so that I do not repeat it. Where did I go wrong here?

user
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patricio2626
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  • Please use MathJax to format your mathematics. As it stands, this is relatively difficult to read. – Arthur Apr 07 '18 at 13:03
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    You can't apply l'Hospital for 2 reasons: 1) it is not a function but a sequence. 2) Also extending to functions $\frac{log x}{-x^{-2}}$ is not an indeterminate form – user Apr 07 '18 at 13:05
  • Thank you Arthur, I will look at this now. I'll accept gimusi's answer when I am able to in 5 minutes. – patricio2626 Apr 07 '18 at 13:09
  • @gimusi can you correct the x under the limit ? It makes no sense – user577215664 Apr 07 '18 at 13:22
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    @Isham yes of course, thanks! – user Apr 07 '18 at 13:23
  • @gimusi "Reason" 1 is ludicrous. Please avoid misleading the OP. – Did Apr 07 '18 at 13:36
  • @Did I did sense that that was incorrect. Convergence seems to intuitively be the limit at infinity. – patricio2626 Apr 07 '18 at 13:38
  • @Did Refer here for details https://math.stackexchange.com/a/1398706/505767 – user Apr 07 '18 at 13:41
  • @gimusi Relevant how? – Did Apr 07 '18 at 13:44
  • @Did Why did you deleted your last comment? Maybe you have finally read the full text and also the comments where is highlighted that there are cases for which $f(n) \to L$ for $n \to \infty$ but $f(x)$ has not limit for $x \to \infty$? Yes maybe it is so and YOU ARE WRONG! since in general l'Hospital can't be extended to sequecences! In many cases we can do but we must be aware about it. Anyway l'Hospital is a bad method (in general) to solve limits if the aim is to understand them and I Always suggest to avoid it whenever possible. – user Apr 07 '18 at 13:49
  • @gimusi You are fantasizing: the suggestion is to deal with the function and to deduce results for the sequence, which always works, not to use the other direction, which indeed may not work. Please read carefully the link you provide. (If reactions from exuberant illiterates similar to yours are to generalize, the site will become difficult to use.) – Did Apr 07 '18 at 13:54
  • @Did But we are not applying l'Hospital rule to the sequence, we are applying the rule to the corresponding function, that is exactly what I've pointed out in my point 2). What are you talking about? You are ludicrous with your attempts to put others in a bad perspective. – user Apr 07 '18 at 13:59
  • @gimusi Stick to the maths if you can: your Reason 1 does not work. Did I even mention Reason 2? No, hence you are either a sophist, or out of your league, or both. – Did Apr 07 '18 at 14:01
  • @Did You are the sophist since extrapolate only one part of my comment without mention point 2 that is a part of it. – user Apr 07 '18 at 14:05
  • @gimusi Quote: "You can't apply l'Hospital for 2 reasons: 1) (something). 2. (something else)." Sorry but, as usual, you are trying to drag this very elementary point into a neverending discussion, so, please... do so if you must, but without me. – Did Apr 07 '18 at 14:08
  • @Did part 2 is not something else since we claim "also extending to functions..." in order to use l'Hospital...etc.. I repeat: l'Hospital can't be applied to the sequences, it can be applied to the corresponding functions, that is exactly what I wrote! Bye – user Apr 07 '18 at 14:15
  • Well, I certainly didn't intend to create this train wreck :-) I do appreciate all of the enthusiasm in the responses to my question. – patricio2626 Apr 07 '18 at 17:29

1 Answers1

2

Note that

$$n^{-n^2}=e^{-n^2\log n}=\frac{1}{e^{n^2\log n}}\to 0$$

indeed

$$n^2\log n\to\infty$$

user
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  • So, is -infinf not an intedterminate form? In other words I should have said limit x-> inf -n^2ln(n) = -inf --> e^-inf --> 1/inf --> 0 ? – patricio2626 Apr 07 '18 at 13:07
  • @patricio2626 yes of course, let me write in symbolic way that $$\large{-\infty \cdot \infty=-\infty}$$ – user Apr 07 '18 at 13:08