Problem: A sequence $(a_n)$ is defined recursively as follows, where $0<\alpha\leqslant 2$:
$$
a_1=\alpha,\quad a_{n+1}=\frac{6(1+a_n)}{7+a_n}.
$$
Prove that this sequence is increasing and bounded above by $2$. What is its limit?
Ideas: How should I go about starting this proof? If the value of $a_1$ were given, I could show numerically and by induction that the sequence is increasing. But no exact value of $\alpha$ is given.
Best Answer
First, $0\leq a_1\leq2$.
Second, $a_{n+1}=\frac{6(1+a_n)}{7+a_n}=6-\frac{36}{7+a_n}$.
So, if $0\leq a_n\leq2$,
$\qquad$then $7\leq 7+a_n\leq 9$,
$\qquad$then $\frac{1}{7}\geq\frac{1}{7+a_n}\geq\frac{1}{9}$,
$\qquad$then $\frac{36}{7}\geq\frac{36}{7+a_n}\geq\frac{36}{9}$,
$\qquad$then $-\frac{36}{7}\leq-\frac{36}{7+a_n}\leq-\frac{36}{9}$,
$\qquad$then $0\leq6-\frac{36}{7}\leq6-\frac{36}{7+a_n}\leq6-\frac{36}{9}=2$,
then $0\leq a_{n+1}\leq2$.
By induction $0\leq a_n\leq2$ for all $n$.
Also $a_{n+1}-a_n=\frac{6(1+a_n)}{7+a_n}-a_n=\frac{6-a_n-a_n^2}{7+a_n}=\frac{(a_n+3)(2-a_n)}{7+a_n}>0$, as long as $0\leq a_n\leq2$.
Therefore $a_{n+1}\geq a_n$.