Continuing With Continued Fractions

The continued fraction is defined as the limit of the sequence

$a_0=3$ ,

$a_1=3-\frac{2}{3}$ ,

$a_2=3-\frac{2}{3-\frac{2}{3}}$ , ...

$a_n=3-\frac{2}{a_{n-1}}$ .

If the sequence converges, say to $a$ , then $a$ satisfies the equation $a=3-\frac{2}{a}$ . Multiplying by $a$ on both sides and solving the resulting quadratic equation shows that if the limit exists, it is either $a=1$ or $a=2$ .

We see that $a_0>2$ . We proceed by induction to show that $a_n>2$ for all $n>0$ , which would then imply that the limit would have to be $a=2$ if it exist.

Assume (by hypothesis) that $a_{n-1}>2$ . Then

$\frac{1}{a_{n-1}}<\frac{1}{2}$

$-\frac{2}{a_{n-1}}>-\frac{2}{2}$

$a_n=3-\frac{2}{a_{n-1}}>3-1=2$

So if the limit exists, it must be 2. To show that the limit exists we show that the sequence is decreasing and bounded below (by 2). To show that the sequence is decreasing, assume that it isn't (and hope for a contradiction). Then there is some $n>0$ so that

$a_n>a_{n-1}$

$3-\frac{2}{a_{n-1}}>a_{n-1}$

$0>a_{n-1}^2-3a_{n-1}+2$

which is true only if $1<a_{n-1}<2$ . This is a contradiction since $a_n>2$ for all $n\geq 0$ . Therefore $\{a_n\}$ is a decreasing sequence bounded below and must converge. We saw above that the limit must be 2.

I did it the way Calvin told me to do i.e $\textbf{OBSERVE}$

QUADRATIC FORMULA.