I want to show that $$ \sum_{k=1}^{n} k^{\alpha} \sim \frac{n^{\alpha+1}}{\alpha+1} \ \ \ \ \ \ n\to \infty$$
for $\alpha > -1$ . So I need to show that the following limit exists and is equal to $1$ : $$ \displaystyle \lim_{n \to \infty} \frac{\sum_{k=1}^{n} k^{\alpha}}{\frac{n^{\alpha+1}}{\alpha+1}} $$
Because $\alpha > -1$, it is clear that the above limit is of the type $\frac{\infty}{\infty}$. Here I have doubts about using the L'Hospital's rule, because that means differentiating with respect to the upper bound of the sum which is a discrete variable, then how shoud I proceed with the computation of this limit?
Thanks,
Use the inequality
$$\int_0^nx^{\alpha}\,dx\leq\sum_{k=1}^nk^{\alpha}\leq\int_1^{n+1}x^{\alpha}\,dx$$
which holds because the middle term is an upper sum for the left integral and a lower sum for the right integral. Equality holds if and only if $\alpha=0$. Now compute both integrals, divide throughout by ${n^{\alpha+1}}$ and let $n\to\infty$.
As pointed out by the8thone, the above holds for $\alpha\geq 0$, and for $-1<\alpha<0$ both inequalities are reversed, but the limits are still the same.