<s>[INST] Prove that $c_n = \frac1n \bigl(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}} \bigr)$ converges I want to show that $c_n$ converges to a value $L$ where:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n}$$
First, it's obvious that $c_n > 0$.
I was able to show using the following method that $c_n$ is bounded:
$$c_n = \frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{n}}}{n} < \overbrace{\frac{\large \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{2}}}{n}}^{n-1 \text{ times}} = \frac{\large {n - 1}}{n\sqrt{2}} < \frac{1}{\sqrt{2}}$$
So now we know that $\large {0 < c_n < \frac{1}{\sqrt{2}}}$.
I know from testing for large values of $n$ that $c_n \to 0$.
What's left is actually finding a way to show this.
Any hints?
 [/INST] We have by the Riemann sum
$$c_n=\frac1n\sum_{k=2}^n\frac1{\sqrt k}=\frac1{n^{3/2}}\sum_{k=2}^n\frac1{\sqrt{\frac kn}}\sim\frac1{\sqrt n}\int_0^1\frac{dx}{\sqrt x}=\frac2{\sqrt n}$$
so clearly $c_n$ tends to $0$ but we found also that
$$\lim_{n\to\infty}c_n\sqrt n=2$$
which we can't find it by applying the Cesàro's theorem.
</s>