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?
 A: Hint : $c_n$ is Cesàro summation of sequence $\{\frac{1}{\sqrt{n}}\}_{n \in \Bbb N}$
Cesàro summation
A: 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.
A: You could try to prove the general, useful fact that
$$\lim_{n\to\infty}a_n=L\implies\lim_{n\to\infty}\frac{a_1+\ldots+a_n}n=L$$
A: Your bracketing $0 < c_n < \frac{1}{\sqrt{2}}$ shows that the sequence is bounded, but is not enough to prove convergence.  Actually, bounding all terms by $1/\sqrt2$ will not lead you to the solution because it is too loose: thanks to the $1/\sqrt n$ decrease, most of the terms are much smaller than that.
Now try to bound above as follows:
$$\frac1{\sqrt2},\frac1{\sqrt3},\frac1{\sqrt4},\frac1{\sqrt5},\frac1{\sqrt6},\frac1{\sqrt7},\frac1{\sqrt8},\frac1{\sqrt9},\frac1{\sqrt{10}},\frac1{\sqrt{11}},\frac1{\sqrt{12}},\frac1{\sqrt{13}},\frac1{\sqrt{14}},\frac1{\sqrt{15}},\frac1{\sqrt{16}}...$$
$$\frac1{\sqrt2},\frac1{\sqrt2},\frac1{\sqrt4},\frac1{\sqrt4},\frac1{\sqrt4},\frac1{\sqrt4},\frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt8},\ \frac1{\sqrt{16}}...$$
(every time the number of equal powers of $2$ doubles; this is the same trick as that used to prove divergence of the harmonic series).
Setting $n=2^m$, what you have now is 
$$\frac1{n}\sum_{k=1}^m\frac{2^k}{\sqrt{2^k}}=\frac1n\sum_{k=1}^m\sqrt{2^k}=\frac1n\frac{(\sqrt2)^{m+1}-\sqrt2}{\sqrt2-1}=\frac1n\frac{\sqrt{2n}-\sqrt2}{\sqrt2-1},$$
by the geometric progression formula. This clearly tends to $0$.
A: 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.
A: 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*}
