hey, i have a question that i cannot figure out.
Given that lim(n->infinity)=an exists when a1=1 and an+1=sqrt[1+(6/an)] for n>1, find the limit.
how do i go about solving this?
cheers!
$\displaystyle a_{n+1}=\sqrt{1+\frac{6}{a_n}}$
$\displaystyle \lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty}\sqrt{1+\frac{6}{a_n}}$
$\displaystyle Let \lim_{n\to \infty}a_{n}=a$,then we also have $\displaystyle \lim_{n\to \infty}a_{n+1}=a$
Therefore,$\displaystyle a=\sqrt{1+\frac{6}{a}}$
$\displaystyle a^2=1+\frac{6}{a}$
$\displaystyle a^2-a-6=0$
$\displaystyle
a=3
$