How to determine the $\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1$. I stuck to do this,

$$\lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}=1.$$

The only thing I have observed is $$ 1\le \lim_{n\to \infty} \frac{1+2^2+\ldots+n^n}{n^n}$$ I am unable to get its upper estimate so that I can apply Sandwich's lemma. 
 A: Consider any sequence $a_n$ with the property that $\lim\limits_{n\to\infty}a_n=\infty$ and $\lim\limits_{n\to\infty}\frac{a_{n-1}}{a_n}=0$. It follows by Stolz–Cesàro that
$$\lim_{n\to\infty}\frac{a_1+a_2+\dots+a_n}{a_n}=\lim_{n\to\infty}\frac{a_n}{a_n-a_{n-1}}=\lim_{n\to\infty}\frac1{1-\frac{a_{n-1}}{a_n}}=1$$
Here, we have $a_n=n^n$.
A: $$1\le \frac{1+2^2+\ldots+n^n}{n^n}\le \frac{n+n^2+\ldots+n^n}{n^n} = \frac{n\frac{n^n-1}{n-1}}{n^n} = \frac{n^{n+1}-n}{n^{n+1}-n^n}\xrightarrow{n\to\infty} 1$$
A: $$1+2^2+3^3+\cdots(n-2)^{n-2}+(n-1)^{n-1}+n^n<\\n^{n-2}+n^{n-2}+n^{n-2}+\cdots n^{n-2}+n^{n-1}+n^n.$$
When divided by $n^n$, these terms are bounded by $\dfrac n{n^2}+\dfrac1n+1$.
A: $$\lim_{n \rightarrow \infty} \frac{1^1 + 2^2 + \dots + (n-1)^{n-1} + n^n}{n^n} =  \lim_{n \rightarrow \infty} \left(\frac{1^1}{n^n} + \frac{2^2}{n^n} + \dots + \frac{(n-1)^{n-1}}{n^n} + 1 \right) $$
Let's examine that second to last term:
$$
\begin{align}
  \lim_{n \rightarrow \infty} \frac{(n-1)^{n-1}}{n^n} &=  \lim_{n \rightarrow \infty} \frac{1}{n}\frac{(n-1)^{n-1}}{n^{n-1}} \\
 &= \lim_{n \rightarrow \infty} \frac{1}{n} \left( \frac{n-1}{n} \right)^{n-1}< \lim_{n \rightarrow \infty} \frac{1}{n} \cdot 1\\
 &= 0
\end{align}$$
The second largest term is zero, none of the terms are negative, the sum is 1:
$$\lim_{n \rightarrow \infty} \frac{\sum\limits_{m=1}^{n} m^m}{n^n} = 1$$
