show the existence of a real number Let $a_n  > 0$ be any sequence. I want to show that there is a real number $r \in \mathbb R$ such that $ 0 < | r - m/n | < a_n $ for infinitely many points $(m,n ) \in \mathbb N ^2$. 
I don't know how to get start. I know that rational numbers are dense in $\mathbb R$, but how can I approximate them to an real number faster than an arbitrary sequence $a_n$? Any help is appreciated.
 A: For a continued fraction approximation, see inequality $(5.4)$, we have
$$
\left|\,r-\frac{p_k}{q_k}\,\right|\le\frac1{q_kq_{k+1}}\tag{1}
$$
and
$$
\begin{bmatrix}p_{k+1}\\q_{k+1}\end{bmatrix}
=c_{k+1}\begin{bmatrix}p_k\\q_k\end{bmatrix}
+\begin{bmatrix}p_{k-1}\\q_{k-1}\end{bmatrix}\tag{2}
$$
If we set the tail of the continued fraction of $r$ to be $(c_k)$ where
$$
c_{k+1}\ge\frac1{q_k^2a_{q_k}}\tag{3}
$$
We can use $(2)$ and $(3)$ to generate a continued fraction for $r$ from the $a_k$.
When $q_{k-1}\gt0$, $(1)$, $(2)$, and $(3)$ imply that
$$
\left|\,r-\frac{p_k}{q_k}\,\right|
\le\frac1{q_kq_{k+1}}
\lt\frac1{c_{k+1}q_k^2}
\le a_{q_k}\tag{4}
$$
A: Let $I_0\subset [0,\infty)$ be a closed interval. Choose $n_1$ such that $1/n_1$ is less than the length of $I_0.$ Then there will be $m_1$ such that $m_1/n_1$ is in the interior of $I_0.$ We can then choose $0<\delta_1 < a_{n_1}$ such that
$$(m_1/n_1-\delta_1, m_1/n_1+\delta_1) \subset I_0.$$
Now choose a closed interval $I_1 \subset (m_1/n_1-\delta_1, m_1/n_1).$
Lather, rinse, repeat. We obtain nested closed intervals $I_1 \supset I_2 \supset \cdots$ and we can conclude $\cap I_k \ne \emptyset.$
Let $r\in \cap I_k.$ Then for each $k,$ we have
$$0< |r-m_k/n_k| < \delta_k < a_{n_k}$$
as desired.
