Show that $a_{n+1}= 1 + \frac{1}{a_n}$ converges. 
Let $\{a_n\}$ be defined by $a_1 =1  $ and $a_{n+1} = 1 + {\dfrac{1}{a_n}}$ with $n \in N$.
Show that $a_{n+1}= 1 + {\dfrac{1}{a_n}}$ converges.

I know the limit but how can I show that is a Cauchy sequence or that this sequence converges.
 A: Hint. Let $f(x)=1+1/x$ then $f:[3/2,2]\to [3/2,2]$ and for $x\in [3/2,2]$,
$$|f(x)-f(y)|= \left|\frac{1}{x}-\frac{1}{y}\right|\leq \frac{4}{9}|x-y|.$$
That is $f$ is a contraction. 
Now $a_{n+1}=f(a_n)$ and $a_2=f(a_1)=2$ and by the Banach fixed-point theorem, $(a_n)_n$ tends to the unique fixed point of $f$ in  the interval $[3/2,2]$.
A: After so many good answers, this one is just for knowledge sharing
$$a_1=1, a_2=2, a_3=\frac{3}{2}, a_4=\frac{5}{3}, ...$$
and by induction
$$a_n=\frac{F_{n+1}}{F_n}$$
where $\{F_n\}$ are Fibonacci numbers, since 
$$a_{n+1}=1+\frac{1}{a_n}=1+\frac{F_n}{F_{n+1}}=\frac{F_{n+1}+F_{n}}{F_{n+1}}=\frac{F_{n+2}}{F_{n+1}}$$
and
$$\lim\limits_{n\rightarrow \infty}\frac{F_{n+1}}{F_{n}}=\varphi=\frac{1+\sqrt{5}}{2}$$
A: Hint: Show that if $a_n < \frac{1 + \sqrt{5}}{2}$ then $a_n < a_{n+2}$.  Similarly, show that if $a_n > \frac{1+\sqrt{5}}{2}$, then $a_n > a_{n+2}$.
A: So let the limit - assuming there is one - be $a$ and note that $2\gt a\gt 1$ and that $a=1+\frac 1a$ whence $a^2=a+1$.
Now write $a_n=a+e_n$ so that $$a+e_{n+1}=1+\frac 1{a+e_n}$$
On clearing fractions we get $$a^2+ae_{n+1}+ae_n+e_ne_{n+1}=a+e_n+1$$ so that $$e_{n+1}=e_n\cdot\frac {1-a}{a+e_n}=-e_n\cdot\frac {a-1}{a+e_n}$$
Use this to show that the error term alternates in sign and decreases in absolute value - you should be able to show that the error reduces geometrically in magnitude, and therefore tends to zero. If the error tends to zero, you have convergence.
This technique of isolating the error may not be the most efficient, but if you are stuck, it can help to show what is going on. Note that all the terms equivalent to the original equation conveniently drop out - this is quite a general phenomenon, and if it doesn't happen, it's an indication either that there is no limit, or that there has been a mistake in calculation.
A: As Marcus M said, first $a_{n+2}=\frac{2a_n+1}{a_n+1}$ showing those properties. Then show $\{a_{2n}\}$ and $\{a_{2n+1}\}$ really convergent to that value.
E.g. for $\{a_{2n}\}$,
because of the fact that $f(x)=\frac{2x+1}{x+1}$ is strictly increasing and  $x>f(x)$ when $x>\frac{\sqrt{5}+1}{2}:=g$. Assume the limit (infimum) is $x^*>g$, for all $x\geqslant x^*$. If There are always  points from $[x^*,x^*+\delta)$. If $\exists\hat{x}, f(\hat{x})\leq x^*$, then $f(f(\hat{x}))=f(x^*)<x^*$ contradict to 
A: Here is yet another answer:  Observe that $a_n \geq 1$ for all $n \in \{1, 2, 3, ...\}$.  Let $b=\frac{1}{2}(1+\sqrt{5})$, which satisfies $b = 1+\frac{1}{b}$ and $b>1$. Then for all $n \in \{1, 2, 3, ...\}$: 
$$ |a_{n+1}-b| = |(1+\frac{1}{a_n}) - (1 + \frac{1}{b})| = \frac{|a_n-b|}{a_nb} \leq \frac{|a_n-b|}{b}$$
where the last inequality uses $a_n\geq 1$.   The error decreases exponentially fast. 
A: Let $f(x) = 1+{1\over x}$. We have $f([1,2]) \subset [1,2]$.
Since $f(1) = 2 > 1$ and $f(2) = {3 \over 2} < 2$ we see that $f$ has
a fixed point in $[1,2]$.
Note that $\phi(x) = f(f(x)) = {2x+1 \over x+1}$ satisfies
$0 \le \phi'(x) \le {1 \over 4}$ for $x \in [1,2]$ hence has
a unique (in $[1,2]$) fixed point $x^*$ and hence $x_{2n+1} \to x^*$
for some $x^* \in [1,2]$. Note that any fixed point of $f$ is a
fixed point of $\phi$, hence $x^*$ is the fixed point of $f$ and
it follows that $x_{2n+2} = f(x_{2n+1})$ converges to $x^*$ as
well.
