Proving divergence of a sequence by proving the sequence is increasing. We define a sequence recursively by $$a_{n+1}=\frac{1}{4}({a_n}^2+a_n+2)~~~~~~~(a_1=3)$$
By showing $a_n$ is increasing prove that $a_n$ does not converge.
Not sure how to do this one. I tried showing that $a_{n+1}-a_n>0$ but couldn't get it to work also not sure how this would be used to imply it doesn't converge. (Although I think you can just state it is unbounded so eventually it will be bigger than any limit you could imagine)
Thanks
 A: rewrite $a_{n+1}=\frac{1}{4}[a_n(a_n+1)]+\frac12$ prove by induction that $a_n+1\ge4$ and $a_{n+1}>a_n\ge3$.
(Do not forget to use that $a_1=3$ for otherwise the proof won't work.)  
Once you do that, if the sequence was convergent then both $a_n$ and $a_{n+1}$ would converge to the same limit $L>3$. You would get 
$L=\frac{1}{4}(L^2+L+2)$. Solve this quadratic equation for $L$ and figure that the roots are too small to be a limit of numbers greater than $3$. 
Alternatively, don't bother with the quadratic equation about $L$, but prove by induction that $a_n+1\ge4$, $a_{n+1}>a_n\ge3$, and $a_{n+1}\ge a_n+\frac12$. Then $a_{n+1} - a_n\ge \frac12$ for all $n$, which in particular implies that the sequence $a_n$ is not Cauchy, and cannot possibly converge. It also shows directly that $a_n\to\infty$, that is, $a_n$ diverges to infinity (which you could have also concluded, indirectly, using that $a_n$ is increasing but has no limit, after you found the possible roots for $L$).  
A: $\begin{array}\\
a_{n+1}
&=\frac{1}{4}({a_n}^2+a_n+2)\\
&=\frac{1}{16}(4{a_n}^2+4a_n+8)\\
&=\frac{1}{16}(4{a_n}^2+4a_n+1+7)\\
&=\frac{1}{16}((2{a_n}+1)^2+7)\\
\end{array}
$
Therefore
$2a_{n+1}+1
=\frac{1}{8}((2{a_n}+1)^2+7)+1
=\frac{1}{8}((2{a_n}+1)^2+15)
$.
Letting
$b_n
=2a_n+1
$,
$b_{n+1}
=\frac{1}{8}(b_n^2+15)
$.
Since $b_1
=2a_1+1
=7
$,
$b_2
=\frac{49+15}{8}
=8
$,
$b_3
=\frac{64+15}{8}
=\frac{79}{8}
> 9
$.
If $b_n
\ge 8+k
$
where $k \ge 1$
(which is true for
$n=3$),
then
$b_{n+1}
=\frac{1}{8}(b_n(8+k)+15)
=\frac{1}{8}(8 b_n+kb_n+15)
=b_n+\frac{1}{8}(kb_n+15)
=b_n+\frac{1}{8}(k(8+k)+15)
$
so
$b_{n+1}-b_n
=\frac{1}{8}(k(8+k)+15)
= k+\frac{1}{8}(k^2+15)
\ge k+2
$.
Therefore
$b_n$
is unbounded
and strictly increasing
and so is 
$a_n$.
