Suppose we have a recursive sequence $a_n$ with $a_1=1$ and $a_{n+1}=1+1/a_n$. Prove that $a_n$ is convergent and determine the limit of the sequence.
So far I have:
First $(a_n)>0,\forall n\in\Bbb N$. For $n>1$, we have $a_{n+1}=1+\frac{1}{a_n}$ and $a_n=1+\frac{1}{a_{n-1}}$.
Note $\max(a_{n+1})=2$. Because $a_2=1+1=2$ and $a_3=1+\frac{1}{2}=\frac{3}{2}$, $a_4=1+\frac{1}{\frac{3}{2}}=\frac{5}{3}$ $a_5=1+\frac{1}{\frac{5}{3}}=\frac{8}{5}$.....
So $\frac{3}{2}\le a_{n+1}-a_n \le 2$, thus,
$$\frac{3}{2}\le a_{n+1}-a_n = \left\lvert\frac{a_{n-1}-a_n}{a_na_{n-1}}\right\rvert\le 2$$ Since $a_n>\frac{3}{2}$
$$ a_{n+1}-a_n= \left\lvert\frac{a_{n-1}-a_n}{a_na_{n-1}}\right\rvert\le2\cdot\left(\frac{2}{3}\right)^2|a_{n-1}-a_n|=\frac{8}{9}|a_{n-1}-a_n|$$
So this is a Cauchy sequence.
Is this correct?