Finding $\lim\limits_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}$ $$\lim_{n \to \infty}{\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}}.$$
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?

Let $f(n) = (1^1 + 2^2 + 3^3 + \cdots + n^n)/n^n$. You want to show $\lim_{n \to \infty} f(n) = 1$.
It's obvious that $f(n) > 1$ for all $n$.
For an upper bound, 
$$ f(n) \le {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} + {n^{n-1} \over n^n} + {n^n \over n^n} = {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} + {1 \over n} + 1.$$
Now I leave it to you to find some bound $g(n)$, with $\lim_{n \to \infty} g(n) = 0$ and
$$ {1^{n-2} + 2^{n-2} + \cdots + (n-2)^{n-2} \over n^n} \le g(n). $$ 
So you have
$$ 1 < f(n) < 1 + {1 \over n} + g(n) $$
and apply the squeeze theorem to finish the proof.
