TITLE: Finding $\lim\limits_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}$
QUESTION [22 upvotes]: $$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$$
With a first look this must give $1$ as a result but have a problem to explain it.
How can I do it?
Edit
I noticed that it is $\frac{\infty}{\infty}$.
$$\lim_{n \to \infty}{n^{n}\frac{(\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1)}{n^n}}= \lim_{n \to \infty}{\frac{1^1}{n^{n}}+\frac{2^2}{n^{n}}+\frac{3^3}{n^{n}}+\cdots+1}=1$$
Is this correct?

REPLY [3 votes]: I know I'm terribly late, but I'd like to post a similar but a bit different from Michael and Greg's approach. Plese tell me in the comments if it's not sufficiently different.
Let $$a_n=\frac{1^1+2^2+3^3+\cdots+(n-1)^{n-1}}{n^n}. $$ So we have $$a_{n+1}=\frac{1^1+2^2+3^3+\cdots+(n-1)^{n-1}+n^n}{(n+1)^{n+1}}=\frac{n^n(a_n+1)}{(n+1)^{n+1}}$$ and thus $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{a_nn^n}{(n+1)^{n+1}}+\frac{n^n}{(n+1)^{n+1}}=\lim_{n\to\infty}\frac{a_nn^n}{(n+1)^{n+1}}.\tag{1}$$ Since $$\lim_{n\to\infty}\frac{n^n}{(n+1)^{n+1}}=0,$$ $(1)$ yields that $a_n$ either converges to $0$ or diverges to $+\infty$. But $$\lim_{n\to\infty}a_n\le\lim_{n\to\infty}\frac{\overbrace{(n-1)^{n-1}+(n-1)^{n-1}+\cdots+(n-1)^{n-1}}^{n-1 \ \text{times}}}{n^n}=\lim_{n\to\infty}\frac{(n-1)^n}{n^n}=e^{-1}, $$ hence $$\lim_{n\to\infty}a_n=0,$$ which allows us to conclude $$\lim_{n\to\infty}\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}=1.$$