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.
 A: Multiplying by $6000$, we have
$$2<\left|6000\sqrt{2}-6000\cdot\frac pq\right|<3.$$
Now, we know that $6000\sqrt2\approx8485.2813742$, so choosing $p,q$ such that $6000p/q=8483$ should do the job. This means $p/q=8483/6000$.
A: Try $$ \frac{58}{41}.$$
It is well-known that the continued fraction of  $\sqrt 2$ is 
$$ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\ldots}}}}$$
Numerically(!), we find the continued fractions for $\sqrt 2+\frac1{2000}$ an $\sqrt 2+\frac1{2000}$:
$$ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{\color{red}1+\frac1{\ldots}}}}}}$$
and
$$ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{\color{red}3+\frac1{\ldots}}}}}.$$This suggests that the simplest fraction inbetween is
$$ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{3}}}}
=1+\frac1{2+\frac1{2+\frac 37}}
=1+\frac1{2+\frac 7{17}}
=1+\frac{17}{41}=\frac{58}{41}.$$
We verify that 
$$ \left(\frac{58}{41}-\sqrt 2\right)\underbrace{\left(\frac{58}{41}+\sqrt 2\right)}_{\approx 2\sqrt 2}=\left(\frac{58}{41}\right)^2-2=\frac2{1681}$$
and hence 
$$ \frac{58}{41}-\sqrt 2\approx \frac1{1681\sqrt 2}$$
which is certainly in the required range.

Edit: On second thought, it turns out that twice the reciprocal of the above, i.e., 
$$ \frac{41}{29}$$
is also a valid (and "simpler") solution, just from below.
A: 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 $\sqrt 2 + \frac\epsilon{2\sqrt 2}$
