<s>[INST] 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. 
 [/INST] $$\lim_{n \rightarrow \infty} \frac{1^1 + 2^2 + \dots + (n-1)^{n-1} + n^n}{n^n} =  \lim_{n \rightarrow \infty} \left(\frac{1^1}{n^n} + \frac{2^2}{n^n} + \dots + \frac{(n-1)^{n-1}}{n^n} + 1 \right) $$
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:
$$\lim_{n \rightarrow \infty} \frac{\sum\limits_{m=1}^{n} m^m}{n^n} = 1$$
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