<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] Just wondering that the following elementary way is not among the presented solutions.
So, I add it here as a late answer:
The following inequality for $k \in \mathbb{N}$ will be used:
$$\sqrt{k+1}-\sqrt k = \frac 1{\sqrt{k+1}+\sqrt k} > \frac 1{2\sqrt{k+1}}$$
Hence,
\begin{eqnarray*} \frac 1n\sum_{k=1}^n \frac 1{\sqrt k}
& = & \frac 1n\left(1+ \sum_{k=1}^{n-1} \frac 1{\sqrt{k+1}}\right) \\
& < & \frac 1n\left(1+ 2\sum_{k=1}^{n-1} (\sqrt{k+1}-\sqrt k)\right) \\
& = & \frac 1n\left(1+ 2(\sqrt{n}-1)\right) \\
& = & -\frac 1n + \frac 2{\sqrt n} \stackrel{n\to\infty}{\longrightarrow} 0
\end{eqnarray*}
</s>