Problem: Find $$\lim_{n\to \infty} \left[\frac{1}{n^k}(1^k+2^k+...+n^k)-\frac{n}{k+1}\right].$$
We solve this problem by taking $x_n=(k+1)(1^k+2^k+...+n^k)-n^{k+1}$ and $y_n=n^k(k+1)$ then by Stolz's theorem we have that $$\lim_{n\to \infty}\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}=\frac{(k+1)n^{k}-(n)^{k+1}+(n-1)^{k+1}}{(k+1)[n^{k}-(n-1)^k]}$$ $$=\frac{(k+1)/n-1+(1-1/n)^{k+1}}{\frac{k+1}{n}[1-(1-1/n)^k]}$$ $$=\frac{(k+1)/n-1+(1-1/n)^{k+1}}{\frac{k+1}{n^2}[(1-1/n)^{k-1}+(1-1/n)^{k-2}+...+1]}$$ $$=\frac{nk+n-n^2+n^2(1-1/n)^{k+1}}{(k+1)[(1-1/n)^{k-1}+(1-1/n)^{k-2}+...+1]}$$
I am not sure how to proceed from here and thus any hints will be much appreciated.
