Find rational $\frac{p}{q}$ such that $\frac{1}{3000}<|\sqrt{2}-\frac{p}{q}|<\frac{1}{2000}$ 
Find rational $\frac{p}{q}$ such that $\frac{1}{3000}<|\sqrt{2}-\frac{p}{q}|<\frac{1}{2000}$

My Attempt
take a sequence which converges to $\sqrt{2}$ : $p_1=1+\frac{1}{2}, p_{n+1}=1+\frac{1}{1+p_n}$
I find how to calculate the sequence : $p_n=\frac{x_n}{y_n}, \Delta y_n=x_n$ and $(\Delta^2-2)y_n=0 $ but there is no way how to find the rational which matches the given term.

You can just do your iteration
$$p_1=\frac 32\\
p_2=1+\frac 1{1+\frac 32}=\frac 75\\
p_3=1+\frac 1{1+\frac 75}=\frac{17}{12}$$
square each, and check how close it is to $2$.  If the square is $2+\epsilon$, the square root is about $\sqrt 2 + \frac\epsilon{2\sqrt 2}$
