TITLE: Approximating $x=\sqrt{2}+1$
QUESTION [7 upvotes]: Suppose $y>1$ is some approximation to $x=\sqrt{2}+1$. Give a brief reason (not a proof) why one should expect $(1/y)+2$ to be a closer approximation to $x$ than $y$ is. 

After testing this out for a bit, it looks like we can let $y_{n+1}=\frac{1}{y_n}+2$ and $\lim_{n\to\infty}y_n=\sqrt{2}+1$, but this does not give me any intuitive idea as to why $y_{n+1}$ should be a better approximation to $x$ than $y_n$ is. 
Can anyone give a brief reason for this improvement in aproximation, especially a more "intuitive" one than simple numerical data?

REPLY [3 votes]: Let$ y<1+\sqrt{2}$, 
So $1+\sqrt{2} - y=D$ (say). 
$1/y > \sqrt{2}-1$
And $1/y+2>1+\sqrt{2}$
So let $1/y+2-(1+\sqrt{2}) =d.$
So we have to prove d < D. 
Let D-d>0. 
Simplifying  you  get :
2$\sqrt{2} >y+1/y $
Which is true as $1-\sqrt{2}<y<1+\sqrt{2}$. 
You can do similarly for $y>1+ \sqrt{2}$