# Series converging to zero?

1. Dec 1, 2009

### fmam3

1. The problem statement, all variables and given/known data
Hi all, I've been banging my head at this one for a while so any help is appreciated!

Consider the series,
$$\frac{1}{n}\sum_{k = 1}^{n} \frac{1}{k^{\delta}}$$
where $$\delta \in (0,1)$$. My teacher tells me that this series converges to zero but I am unable to prove that it is so.

2. Relevant equations

3. The attempt at a solution
Indeed, I can easily show that this series converges (i.e. to something, but not necessarily to zero) by doing the following. Since for $$k = 1, 2, \ldots, n$$, we have that $$k \leq n$$, then it folloWs that $$k^{\delta} \cdot k = k^{1 + \delta} \leq k^{\delta}n$$, and this implies we have $$\frac{1}{k^{1 + \delta}} \geq \frac{1}{k^{\delta}n}$$. And since $$1 + \delta > 1$$, then it implies $$\sum \frac{1}{k^{1 + \delta}}$$ is a convergent p-series and hence by the Comparision Test, it follows that $$\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n} \frac{1}{k^{\delta}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k = 1}^{n} \frac{1}{k^{\delta}} < \infty$$.

--- but then again, what I need is to show that the series not only converges, but converges to zero! (note that if we had $$1/n^2$$, instead of just $$1/n$$, then this problem immediately follows from what I'd shown above. Thanks!