Approximation to $ \sqrt{2}$ I'm a first year Undergraduate student from India. Our professor is going to start a Real Analysis course in September and I was preparing for the initials. I tried and solved many problems, but this one has me confused. Probably the main reason for the confusion is that my book has cited it as Hardy's problem.

If $\dfrac {m}{n}$ is a good approximation to $\sqrt{2}$, prove that $\dfrac{m+2n}{m+n}$ is a better one, and that the errors in the two cases are in opposite direction. Apply this result to show that the limit of the sequence $\dfrac{1}{1}$, $\dfrac{3}{2}$,$\dfrac{7}{5}$,$\dfrac{17}{12}$,$\dfrac{41}{29}$,.... is $ \sqrt{2}$.

I need help regarding the first part of the problem, since the second part is obvious. The simpler the language, the better it is for me.
 A: We do the "opposite directions" and "better approximation" parts, including an estimate of how much better.  
We are intended to assume that $m$ and $n$ are positive, and indeed that they are positive integers. In the argument below, we do  not need $m$ and $n$ to be integers, but we do assume they are positive.  Some assumption needs to be made, since $m=-1$, $n=1$ quickly leads to disaster!
Look at 
$$\frac{m+2n}{m+n}-\sqrt{2}.$$
This is equal to
$$\frac{m+2n-m\sqrt{2}-n\sqrt{2}}{m+n},$$
which in turn is equal to
$$-\frac{(\sqrt{2}-1)(m-n\sqrt{2})}{m+n}.$$
Divide top and bottom by $n$. We get that the above expression is equal to
$$-\frac{(\sqrt{2}-1)(\frac{m}{n}-\sqrt{2})}{1+\frac{m}{n}}.$$
So we conclude that 
$$\frac{m+2n}{m+n}-\sqrt{2}=\left(-\frac{\sqrt{2}-1}{1+\frac{m}{n}}\right)\left(\frac{m}{n}-\sqrt{2}\right).$$
Note that the "multiplication factor" $-\frac{\sqrt{2}-1}{1+\frac{m}{n}}$ is negative.  That means that if $\frac{m}{n}-\sqrt{2}$ is negative, then $\frac{m+2n}{m+n}-\sqrt{2}$ is positive, and if $\frac{m}{n}-\sqrt{2}$ is positive, then $\frac{m+2n}{m+n}-\sqrt{2}$ is negative. Thus the approximations alternate between too big and too small.
Note also that the multiplication factor $-\frac{\sqrt{2}-1}{1+\frac{m}{n}}$ has absolute value less than $\sqrt{2}-1$, which is less than $0.5$. So the absolute value of the error when we approximate $\sqrt{2}$ by $\frac{m+2n}{m+n}$ is less than half the absolute value of the error when we approximate $\sqrt{2}$ by $\frac{m}{n}$.  
Note that we can make a better estimate of the rate of approach to $\sqrt{2}$, if we assume that we start with $m=n=1$. For then, forever, our approximation will be bigger than $1$, so the multiplication factor has absolute value $(\sqrt{2}-1)/(1+m/n)$, which is less than $(\sqrt{2}-1)/2$.
A: First, let $x_k = \frac{m}{n}$, then $x_{k+1} = \frac{x_k +2}{x_k+1}$. 
Notice that  $x_{k+1} - x_k = \frac{2-x_k^2}{1+x_k}$ and that $x_{k+1}^2 - 2 = \frac{2-x_k^2}{(1+x_k)^2}$. 
Thus if $0< x_k <\sqrt{2}$, then $x_{k+1} > \sqrt{2}$. Also from here 
$$
   \vert x_{k+1}^2 - 2 \vert < \vert x_k^2 - 2 \vert
$$
for $x_k > 0$. Thus $x_k$ converges to $\sqrt{2}$ and $x_k - \sqrt{2}$ is an alternating sequence.
A: Suppose $\frac{m}{n} $ is slightly bigger than $\sqrt{2}$, so that we can write $\frac{m}{n}= \sqrt{2}(1+\epsilon)$ where $\epsilon >0$ is small. 
Then $$\frac{m+2n}{m+n} = \frac{ \frac{m}{n} +2}{\frac{m}{n} +1} = \frac{ \sqrt{2}(1+\epsilon)+2}{\sqrt{2}(1+\epsilon)  + 1} = \sqrt{2} \left(1- \left(\frac{\sqrt{2}-1}{\sqrt{2}+1+\sqrt{2}\epsilon}\right)\epsilon \right)$$
Note that $\sqrt{2}+1+\sqrt{2}\epsilon> \sqrt{2}+1 $. Also, since $1<\sqrt{2}< \frac{3}{2}$,  we have $\frac{\sqrt{2}-1}{\sqrt{2}+1} < \frac{1}{4}$ so ,$$\frac{\sqrt{2}-1}{\sqrt{2}+1+\sqrt{2}\epsilon}<\frac{1}{4}.$$
Thus, $\frac{m+2n}{m+n}$ is slightly smaller than $\sqrt{2}$ and it's difference from $\sqrt{2}$ is smaller in magnitude than the previous estimate, and decreases by at least a factor of $4$ with each iteration. 
In a similar manner you can show the other case.
EDIT: I strengthened the estimates to address the rate of convergence issues Andres Caicedo brought up in a comment above. 
A: HINT $\ $ Note that since $\rm\:\dfrac{m}n\:\dfrac{2\:n}m\:=\:2,\:$ one fraction is less than $\:\sqrt{2}\:$ and the other greater. Further their mediant $\rm\:\dfrac{m+2n}{n+m}\:$ is strictly between them, being the slope of the diagonal $\rm\:(n,m)+(m,2n)\:$ of the parallelogram formed by the vectors $\rm\:(n,m)\:$ and $\rm\:(m,2n)\:.\:$ To learn more search on the terms: mediant, Farey series and continued fraction.
A: A somewhat surreal and lazy solution using Mathematica to make things easier: 
(* your two guesses *) 
guess1 = m/n 
guess2 = (m+2*n)/(m+n) 

(* if you square your guesses and subtract from 2, you get signed 
closeness to 2; squaring again eliminates the sign *) 

dist1 = (guess1^2-2)^2 
(-2 + m^2/n^2)^2 

dist2 = (guess2^2-2)^2 
(m^2 - 2*n^2)^2/(m + n)^4 

(* We want dist2 < dist1; under what cases can that fail? *) 
Reduce[{dist1 <= dist2, m>0, n>0}, Reals] 
n > 0 && m == Sqrt[2]*Sqrt[n^2] 

So, the only case where dist1 is even EQUAL to dist2 is when m/n is 
Sqrt[2], which is, of course, impossible. 
