Convergence rate of a series What is convergence rate of a series
$  
\sum_{k=1}^{n} k^\alpha
\\ $
where $\alpha< -1$ ? For example, for $\alpha=-1$ it equals to $O(\log n)$.
 A: Edit: This answer was wrong. I posted a new answer, and I'm keeping this for the comments.
A: My previous answer contains a mistake. The difference between the series and the integral is actually not negligible.
First of all, it is correct to say that $\sum_{k=1}^n k^\alpha = O(1)$, because as $n \to \infty$, the series converges when $\alpha < -1$.
If you want to know a more precise asymptotic formula, you can use the following: First of all, note that $\sum_{k=1}^\infty k^\alpha = \zeta(-\alpha)$, where $\zeta$ is the Riemann zeta function. Then use the fact that
$$\sum_{k=1}^n k^\alpha = \sum_{k=1}^\infty k^\alpha - \sum_{k=n+1}^\infty k^\alpha = \zeta(-\alpha) - \sum_{k=n+1}^\infty k^\alpha$$
so our task is to find the rate of decay of $\sum_{k=n+1}^\infty k^\alpha$. One simple way to do this is to compare this sum to the integral $\int_{k=n}^\infty x^\alpha dx$. This graph should explain why this works: The sum gives you the blue region, while the integral gives you the green region (in the graph, $n=1$).

So, we have:
$$0 \le \sum _{k=n+1}^\infty k^\alpha \le \int_{k=n}^\infty x^\alpha dx = - \frac {n^{\alpha + 1}} {\alpha + 1}$$
This means:
$$\sum_{k=1}^n k^\alpha = \zeta(-\alpha) + O\left(n^{\alpha + 1}\right)$$
You can get further terms by using Abel's summation formula to convert the sum $\sum_{k=n+1}^\infty k^\alpha$ to an integral, and repeat as much as you want (and can, based on $\alpha$) using integration by parts.
