1. ## Help with limit

I need help finding the limit of

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

2. n is natural, and n->+inf.
The problem is you can't separate the expression and calculate the product of both limits...

3. Hello, gunt!

I assume that $n \to \infty$

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

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

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

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

$\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}
$

5. 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

6. According to Maple,

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

Not 1.

7. Originally Posted by AfterShock
According to Maple,

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

Not 1.
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 $\pm \infty$ 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 $x$, but that does not succesfully express the outer function.

8. 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...

9. Originally Posted by AfterShock
According to Maple, lim(as x approaches infinity) [((1+(1/x))^x)^x]/(e^x) = 1/sqrt(e) Not 1.
But try as I have, I cannot justify it.

10. Originally Posted by topsquark
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.
I understand that this problem was posted and answered somewhere else too? I don't know TPH's solution, but I think I've explained why Soroban's isn't correct. The answer definitely isn't 1, but exp(-1/2).

11. Originally Posted by TD!
I understand that this problem was posted and answered somewhere else too? I don't know TPH's solution, but I think I've explained why Soroban's isn't correct. The answer definitely isn't 1, but exp(-1/2).
Did I do this right?

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

By a Taylor expansion (since 1/n is so small when we take the limit) we get that:
$ln \left ( 1 + \frac{1}{n} \right ) = \frac{1}{n} - \frac{1}{2n^2} +...$

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

So for the problem:
$\lim_{n \to \infty}e^{-n} \left ( 1 + \frac{1}{n} \right )^{n^2}$ = $\lim_{n \to \infty}e^{-n} e^{ \left ( n - \frac{1}{2} +... \right ) }$ = $\lim_{n \to \infty}1 \cdot e^{-1/2} = e^{-1/2}$

-Dan

12. Great! Thanks for your help.

13. Originally Posted by topsquark
Did I do this right?

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

By a Taylor expansion (since 1/n is so small when we take the limit) we get that:
$ln \left ( 1 + \frac{1}{n} \right ) = \frac{1}{n} - \frac{1}{2n^2} +...$

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

So for the problem:
$\lim_{n \to \infty}e^{-n} \left ( 1 + \frac{1}{n} \right )^{n^2}$ = $\lim_{n \to \infty}e^{-n} e^{ \left ( n - \frac{1}{2} +... \right ) }$ = $\lim_{n \to \infty}1 \cdot e^{-1/2} = e^{-1/2}$

-Dan
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.

14. Originally Posted by ThePerfectHacker
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.
I was wondering about the same point myself. Unfortunately (and you can blame this on my being a Physicist) I have a fairly poor knowledge of how to work with infinite series, so I can't even begin to figure out how to approach that question.

-Dan

15. Originally Posted by topsquark
I was wondering about the same point myself. Unfortunately (and you can blame this on my being a Physicist) I have a fairly poor knowledge of how to work with infinite series, so I can't even begin to figure out how to approach that question.
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.

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