If $a_1=1$, $a_{n+1}=(a_n+1/4)^2$, then the sequence diverges to $\infty$? 
If $a_1=1$, $a_{n+1}=(a_n+1/4)^2$, then the sequence diverges to $\infty$?

So I've shown that the sequence is monotone increasing (if it's relevant), and my proof goes like:
Let $\epsilon>0$, and $N=$ "something", then for every $n>N$, $a_n=(a_{n-1}+1/4)^2 >$ "some kind of function of $n >$ some kind of function of $N =\epsilon$.
I haven't seen any examples of such proof where the sequence is defined recursively, so I really don't know how to fill in the gaps.
 A: Consider the difference between terms, 
$$a_{n+1} - a_n = (a_n+1/4)^2 - a_n  = (a_n - 1/4)^2 \ge 0$$
For the starting point $a_1 = 1$, all term differences are positive:
$$a_{n+1} > a_n > \cdots > a_2 > a_1$$
And for larger $a_n$, the difference is larger. For any $n$,
$$\begin{align*}
a_{n+1} - a_n &= (a_n-1/4)^2\\
&\ge (a_1-1/4)^2\\
&= (3/4)^2
\end{align*}$$
So by telescoping,
$$a_{n+1} -a_1 \ge n(3/4)^2$$
So the sequence is monotonically increasing and unbounded.
A: Note that $a_n \ge n$ for $n > 3$. 
So $\forall \epsilon > 0$, $\exists N=\max(4, \lceil{\epsilon}\rceil)$, s.t. $a_N > \epsilon$,
A: As $(a_n)_n$ is monotonic, it is convergent if and only if bounded. Now, if $(a_n)_n$ converges, its limit $a$ satisfies $a=(a+1/4)^2$, i.e. $a=1/4$. This is a contradiction because $a_n \geq a_0 >1>1/4$ for every $n$.
A: Simple.  We know that
$$a_1=1\implies a_2=\left(\frac54\right)^2$$
And that
$$a_{n+1}=\left(a_n+\frac14\right)^2>a_n^2$$
Thus, for $n>1$, we have
$$a_n>\left(\frac54\right)^{2^{n-1}}\gg\left(\frac54\right)^n\to\infty$$
A: Assume the sequence converges to some real number $l$. Then we have $l=l^2+1/16+l/2$ or $16l^2-8l+1=0$, whence $l=1/4$. But since $a_1=1$ and the sequence is increasing we have for each $n\in\mathbb{N},a_n\geq 1$. So $1/4=l=\lim a_n\geq 1$ which is a contradiction.
