Does This Mathematical Series Converge to Zero?

[fmam3]

Homework Statement

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.

Homework Equations

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!

 

[mighty2000]

Thank you for bringing this interesting problem to our attention. I can see that you have already made some progress in showing that the series converges, but as you mentioned, the key is to show that it converges to zero. Let me offer some additional insights and suggestions that may help you in your proof.

Firstly, we can rewrite the series as follows:

\frac{1}{n}\sum_{k = 1}^{n} \frac{1}{k^{\delta}} = \frac{1}{n} \cdot \frac{1}{1^{\delta}} + \frac{1}{n} \cdot \frac{1}{2^{\delta}} + \frac{1}{n} \cdot \frac{1}{3^{\delta}} + \cdots + \frac{1}{n} \cdot \frac{1}{n^{\delta}}

Notice that each term in the series is of the form \frac{1}{n} \cdot \frac{1}{k^{\delta}} where k ranges from 1 to n. This means that as n approaches infinity, the terms in the series become smaller and smaller, since the denominator increases while the numerator stays constant. In other words, the terms in the series approach zero as n approaches infinity.

Now, to show that the series converges to zero, we need to show that the sum of the terms also approaches zero as n approaches infinity. To do this, we can use the comparison test again. We can compare the terms in this series to the terms in a convergent p-series with p > 1, such as \frac{1}{n^2}. Since \delta \in (0,1), we can choose p such that p > \frac{1}{\delta}. This means that \frac{1}{n^p} will decrease faster than \frac{1}{n^{\delta}} as n approaches infinity, and therefore, the series \sum \frac{1}{n^p} will converge faster than the series \sum \frac{1}{n^{\delta}}.

Since the series \sum \frac{1}{n^p} converges, by the comparison test, the series \sum \frac{1}{n^{\delta}} must also converge. And since we have already shown that the terms in the series approach zero as n approaches infinity, we