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?
 A: 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 $$\lim_{n\to\infty}\frac{1^1+2^2+3^3+\cdots+n^n}{n^n}=1.$$
A: We can write $(1^1+2^2+\cdots+n^n)/n^n$ as $a_n + b_n + 1$, where
$$
a_n = \frac{1^1+2^2+\cdots+(n-2)^{n-2}}{n^n} \text{ and } b_n = \frac{(n-1)^{n-1}}{n^n}.
$$
Both $a_n$ and $b_n$ are positive, and also
$$
a_n < \frac{(n-2)(n-2)^{n-2}}{n^n} < b_n < \frac{n^{n-1}}{n^n} = \frac1n.
$$
The squeeze theorem should allow you to prove that your answer of 1 is correct.
A: Stolz-Cesàro is probably overkill, but solves the problem easily.
The limit 
$$\lim\limits_{n\to\infty} \frac{(n+1)^{n+1}}{(n+1)^{n+1}-n^n} =1 \,,$$
is very simple to calculate.
A: 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.
