Example: We try to determine the value of $$\lim_{n\to \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}},\,k\in \mathbb{N}.$$ We consider the sequences $\alpha_{n\geq 1}=1^k+2^k+...+n^k$ and $\beta_{n\geq 1}=n^k$ and using the Stolz-Cesaro theorem we have that \begin{eqnarray} \lim_{n\to \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\\ \lim_{n\to \infty}\frac{(1^k+2^k+...+(n+1)^k)-(1^k+2^k+...+n^k)}{(n+1)^{k+1}-n^{k+1}}=\\ \lim_{n\to \infty}\frac{(n+1)^k}{{(n+1)^{k+1}-n^{k+1}}}. \end{eqnarray}
Now we try to get the expression in the indeterminate form $\frac{0}{0}$ as n approaches $\infty$ dividing numerator and denominator of (3) by $(n+1)^k$ \begin{eqnarray} \lim_{n\to \infty}\frac{1}{(n+1)-n^{k+1}(n+1)^{-k}}=\\ \lim_{n\to \infty}\frac{1}{n(1+n^{-1}-(1+n^{-1})^{-k})}=\\ \lim_{n\to \infty}\frac{n^{-1}}{1+n^{-1}-(1+n^{-1})^{-k}}. \end{eqnarray}By applying L'Hôpital's rule once we get \begin{eqnarray} \lim_{n\to \infty}\frac{n^{-1}}{1+n^{-1}-(1+n^{-1})^{-k}}=\\ \lim_{n\to \infty}\frac{-n^{-2}}{-n^{-2}-k(1 +n^{-1})^{-k-1}n^{-2}}=\\ \lim_{n\to \infty}\frac{1}{1+k(1+n^{-1})^{-k-1}}=\\ \frac{1}{1+k}. \end{eqnarray}