I need help finding the limit of

e^(-n) * (1 + 1/n)^(n^2)

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- Oct 28th 2006, 08:41 AMguntHelp with limit
I need help finding the limit of

e^(-n) * (1 + 1/n)^(n^2) - Oct 28th 2006, 09:47 AMgunt
n is natural, and n->+inf.

The problem is you can't separate the expression and calculate the product of both limits... - Oct 28th 2006, 01:29 PMSoroban
Hello, gunt!

I assume that $\displaystyle n \to \infty$

Quote:

$\displaystyle \lim_{n\to\infty}\left[e^{-n}\cdot\left(1 + \frac{1}{n}\right)^{n^2}\right]$

We have: .$\displaystyle \frac{\left(1 + \frac{1}{n}\right)^{n^2}}{e^n} \;= \;\frac{\left(1 + \frac{1}{n}\right)^{n\cdot n}}{e^n} \;=\;\frac{\left[\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} $

Then: .$\displaystyle \lim_{n\to\infty} \frac{\left[\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} \;= \;\frac{\left[\lim_{_{n\to\infty}}\left(1 + \frac{1}{n}\right)^{n}\right]^n}{e^n} \;=\;\frac{e^n}{e^n} \;=\;1 $

- Oct 28th 2006, 02:01 PMTD!
This can't be correct:

- the denominator depends on 'n' as well so you can't just shift the limit to the numerator only.

- in the numerator, you ignored the top exponent n, applied the standard limit and then just put the exponent back.

$\displaystyle \begin{array}{*{20}l}

{\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)^{n^2 } = \mathop {\lim }\limits_{n \to \infty } \left( {\left( {1 + \frac{1}{n}} \right)^n } \right)^n } & { \ne \left( {\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right)^n } \right)^n = e^n } \\

{} & { = \mathop {\lim }\limits_{n \to \infty } \left( e \right)^n = + \infty } \\

\end{array}

$ - Oct 28th 2006, 03:55 PMtopsquark
Hmmmmm....

That's the same "error" that TD pointed out (if error it is). And it appears I lied... The expression with n = 1000 (as far as my calculator will go) appears to be decreasing but fairly flat (first derivative at n = 1000 is -2 x 10^(-7)) and only at 0.6 or so.

I can't say why TPH and Soroban's methods would be wrong, but I don't think the answer is 1.

-Dan - Oct 28th 2006, 04:17 PMAfterShock
According to Maple,

lim(as x approaches infinity) [((1+(1/x))^x)^x]/(e^x) = 1/sqrt(e)

Not 1. - Oct 28th 2006, 04:52 PMThePerfectHacker
You seem to be right, I graphed it too.

However, I am confused why my results produced the incorrect answer. It is certainly that the limit composition rule failed, for some reason. Perhaps, when I said it works for $\displaystyle \pm \infty$ was wrong, I know that it works for numbers, I just always assumed it works for infinite limits also. Apparently it seems by assumption was wrong.

Though I can prepared to be wrong, and think I am. What you shown was not a proof :). It needs to be formal.

EDIT. I realized my mistake.

What I said about composition functions holds true. But I have not properly expressed the outer function, I assumed it was raised to the $\displaystyle x$, but that does not succesfully express the outer function. - Oct 28th 2006, 05:44 PMgunt
I tried a few algebric manipulations and got stuck on each one of them.

It's giving me more trouble than I expected. It looked so innocent... - Oct 28th 2006, 06:47 PMPlato
- Oct 29th 2006, 03:32 AMTD!
- Oct 29th 2006, 05:04 AMtopsquark
Did I do this right?

$\displaystyle \left ( 1 + \frac{1}{n} \right ) ^{n^2} = e^{n^2 \cdot ln \left ( 1 + \frac{1}{n} \right ) }$

By a Taylor expansion (since 1/n is so small when we take the limit) we get that:

$\displaystyle ln \left ( 1 + \frac{1}{n} \right ) = \frac{1}{n} - \frac{1}{2n^2} +...$

So:

$\displaystyle \left ( 1 + \frac{1}{n} \right ) ^{n^2} = e^{n^2 \cdot ln \left ( 1 + \frac{1}{n} \right ) } = e^{n^2 \cdot \left ( \frac{1}{n} - \frac{1}{2n^2} +... \right ) }$ = $\displaystyle e^{ \left ( n - \frac{1}{2} +... \right ) }$

So for the problem:

$\displaystyle \lim_{n \to \infty}e^{-n} \left ( 1 + \frac{1}{n} \right )^{n^2}$ = $\displaystyle \lim_{n \to \infty}e^{-n} e^{ \left ( n - \frac{1}{2} +... \right ) }$ = $\displaystyle \lim_{n \to \infty}1 \cdot e^{-1/2} = e^{-1/2}$

-Dan - Oct 29th 2006, 05:21 AMgunt
Great! Thanks for your help.

- Oct 29th 2006, 05:26 AMThePerfectHacker
I was also thinking about using a series expansion on this one, but I think you made it as simple as possible. Good job.

It looks correct to me. Again as I said the limit composition rule is used here.

The only think I am concerned about is whether a distribution of the limit to each individual term is allowed in an infinite series. I think yes, I believe you can always do it. Or whether the series need to be absolutely convergent, but I think you are correct about this one. - Oct 29th 2006, 06:06 AMtopsquark
- Oct 29th 2006, 06:12 AMThePerfectHacker
Not to offend you but I realized that in my engineering class. The professor just manipulates that stuff and differencials in an way he wants to.

It some ways I agree with that approach. I think I am the only mathemation in my differencial equations class. I have made some remarks about the professors' solution after class, he told me he would be concered with it if he was in a roomful of mathemations but engineers do not need to know that stuff. In fact, If I want to for example, differencial equations on a serious level I would first learn how to solve them informally and incorrectly. Then I would learn the theory, so that I not have to memorize techniques to how to solve them, that stuff I would already know and only be concerned with about the theory.