# How to show that $\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^{n^2}=\infty$

Let's say I need to find the limit: $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n^2}$$ So I know that the limit is $\infty$, but I'm not sure how to show it in situations like this. $$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n^2}=\lim_{n\to\infty}\left(\left(1+\frac{1}{n}\right)^{n}\right)^n=e^{\infty}=\infty$$ I'm pretty sure this one is wrong, because I can't just have the power of infinity. How do I write it down properly?

Also, is $\lim_{n\to\infty}1^n=1$? I know I can't just write $\lim_{n\to\infty}1^n=1^{\infty}$, how should I do it properly?

• I'd accept that just fine. However, you could take the $\log$ then do the limit then re-exponentiate to get the solution. – mathematics2x2life Dec 3 '13 at 19:31
• Re the question at the end: $\lim_{n\to\infty}1^n=\lim_{n\to\infty}1=1$. – Carsten S Dec 3 '13 at 19:42

## 3 Answers

Recall that for $a > 0$ and $m > 1$, we have $$(1+a)^m > 1+ma \tag{\star}$$ The proof for $(\star)$ follows immediately from Binomial theorem. Hence, in your case, we have $$\left(1+\dfrac1n\right)^{n^2} > 1 + \dfrac{n^2}n = 1 +n$$ Now conclude what you want.

Hint:
Since $\lim\left(1+\frac1n\right)^n=e>2$ it follows that for large $n, \ \left(1+\frac1n\right)^{n^2}>2^n$.

If you know the limit of $(1+\frac 1n)^n$ and how to derive it, you ought to be able to show that if $n$ is large enough, $(1+\frac 1n)^n\gt a \gt 1$ so for $n$ large enough $(1+\frac 1n)^{n^2}=\left((1+\frac 1n)^n\right)^n\gt a^n$

I'll leave the gaps for you to fill in.