calculus - A familiar limit, but in general form .

Suppose $a \in \mathbb{R} ,a>1$
Is there an idea to compute the limit below ?

$$\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)$$ I tried it for $a=1,2,3$ but I get stuck in general form . Can someone help me ? Thanks in advance.

Answer

By Stolz theorem (see here: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) we obtain:

$$\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)=$$
$$=\frac{1}{a+1}\lim_{n\rightarrow+\infty}\frac{(a+1)(1^a+2^a+3^a+...+n^a)-n^{a+1}}{n^a}=$$

$$=\frac{1}{a+1}\lim_{n\rightarrow+\infty}\frac{(a+1)n^a-n^{a+1}+(n-1)^{a+1}}{n^a-(n-1)^a}=$$
$$=\frac{1}{a+1}\lim_{n\rightarrow+\infty}\frac{(a+1)n^a-n^{a+1}+n^{a+1}-(a+1)n^a+\frac{a(a+1)}{2}n^{a-1}+...}{n^a-n^a+an^{a-1}+...}=\frac{1}{2}.$$