How to calculate $\lim_{n \rightarrow \infty} e^{-n} (\frac{1}{n}+1)^{n^2}$?

Question

How to calculate $$\lim_{n \rightarrow \infty} e^{-n} (\frac{1}{n}+1)^{n^2}$$?

The result should be $$\frac{1}{\sqrt{e}}$$

But I have no idea how to get it.

I've been only thinking of rules to multiply the $e^{-n}$ through somehow, but I don't seem to find anything. The different quantities also seem to work in different directions making interpretation difficult, since: $$e^{-n}\rightarrow 0$$ $$(\frac{1}{n}+1) \rightarrow 1$$ $$n^2 \rightarrow \infty$$

So these interactions make it difficult to say anything about this.

Perhaps there's a way to develop inequalities that show which terms will dominate when approaching infinity?

Answer 1

Hints: some ideas with Taylor expansions as the comment says. Put

$$a_n=e^{-n}\left(1+\frac1n\right)^{n^2}\implies \log a_n=-n+n^2\log\left(1+\frac1n\right)=$$

$$=-n+n^2\left(\frac1n-\frac1{2n^2}+\mathcal O\left(\frac1{n^3}\right)\right)=\ldots$$

Answer 2

If we take logarithm, we get

$$-n+n^2log(1+\frac{1}{n})$$

$$=-n+n^2(\frac{1}{n}-\frac{1}{2n^2}(1+\epsilon(n))$$

with $\epsilon(n)$ going to $0$ when

$n \to +\infty$.

thus

the limit is

$$e^{-\frac{1}{2}}.$$