Let $u_1$=2 and $u_{n+1}=2+\frac{1}{u_n}$ for $n\geq 1$. Prove that the sequence converges to $\sqrt{2} +1$.
I have no clue how to do it,probably using Cauchy sequence definition. The hint given in textbook is: $|u_{n+2}-u_{n+1}|<\frac{1}{4}|u_{n+1}-u_n|$
Any help is appreciated!
First, for $n>1$, $u_n > 2$ so $u_n < 5/2$ so it is bounded.
Then
$u_{n+2}-u_{n+1} =(2+1/u_{n+1})-(2+1/u_{n}) =1/u_{n+1})-1/u_{n} =(u_n-u_{n+1})/(u_nu_{n+1}) $
and
$u_nu_{n+1} > 4$.
Take absolute values to get the result.
The standard next step is to iterate this to get
$|u_{n+k+1}-u_{n+k}| < |u_{n+1}-u_n|/4^k$
and then use the triangle inequality to show that $|u_{n+k}-u_n| \to 0$ as $n \to \infty$ for all $k$.
This works for
$|u_{n+2}-u_{n+1}| < |u_{n+1}-u_{n}|/(1+c)$
for any $c>0$.
