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

$$1\le \frac{1+2^2+\ldots+n^n}{n^n}\le \frac{n+n^2+\ldots+n^n}{n^n} = \frac{n\frac{n^n-1}{n-1}}{n^n} = \frac{n^{n+1}-n}{n^{n+1}-n^n}\xrightarrow{n\to\infty} 1$$

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 $a_n=n^n$.