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?

We have $c_n>0$ and
$$\begin{align}c_{n+1}-c_n&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-\cdots-\frac{1}{\sqrt{n}}\right]\\
&<\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n}}-\cdots-\frac{1}{\sqrt{n}}\right]\\
&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{n-1}{\sqrt{n}}\right]\\
&<0\end{align}$$
so $c_{n+1}<c_n$ and convergence follows from monotone convergence theorem.
