Prove or evaluate that:$$ \lim_{ n \to \infty }\frac{S_{k}(n)}{n^{k+1}}=\frac{1}{k+1}$$Where$$S_{n}^k=\sum^{n}_{m=0}m^k$$
So I have noticed that it can be done by proving $S_k(n)$ is a polynomial of degree $n+1$ while the leading coefficient is $\frac{1}{k+1}$, but how do I expand it? Or, if this method is not recommended, can you show me how to prove it?
And here is my incomplete attempt:
$$\sum^n_{m=0}m^k=\sum^{n}_{m=0}{(n-m)^k}=\sum^{n}_{m=0}(n-(n-m))^k=\sum^{n}_{m=0}\sum^{k}_{i=0}\binom{k}{i}n^{k-i}(m-n)^i$$$$=\sum^{n}_{m=0} \sum^{k}_{i=0} \sum^{i}_{j=0}\binom{k}{i}\binom{i}{j}n^{k-i}m^{j}{(-n)}^{i-j}$$
Edit:Since the sequence in the denominator is strictly monotone and tending to infinity, we can use stolz theorem.
Using stolz theorem, we get\begin{align}\lim_{n\to \infty}\frac{S_k(n)}{n^{k+1}}=\\\lim_{n\to \infty}\frac{S_k({n+1})-S_k(n)}{(n+1)^{k+1}-n^{k+1}}=\\\lim_{n\to \infty}\frac{(n+1)^k}{(n+1)^{k+1}-n^{k+1}}=\\\lim_{n\to\infty}\frac{\sum_{i=0}^{k}\binom{k}{i}n^i}{\sum_{i=0}^{k}{\binom{k+1}{i}}n^i}=\\\lim_{n\to\infty}\frac{n^k+\cdots}{(k+1)n^k+\cdots}=\\\frac{1}{k+1}\\\end{align}