Define a sequence by

$\displaystyle a_{n+1}=\frac{a_n+1}{2}$, $\displaystyle n\ge 0$

Prove that if $\displaystyle a_0\le 1$, the sequence is increasing and bounded above.

My solution:

$\displaystyle a_{n+1}-1=\frac{a_n+1}{2}-1=\frac{a_n-1}{2}$

So, $\displaystyle a_0\le 1\implies a_n\le 1$ for all $\displaystyle n$, ie the sequence is bounded above by 1.

$\displaystyle a_{n+1}-a_n=\frac{a_n+1}{2}-a_n$

$\displaystyle =\frac{1-a_n}{2}$

We know that $\displaystyle a_n\le 1$ for all $\displaystyle n$. So, $\displaystyle a_{n+1}\ge a_n$ and the sequence is increasing.