Proving that $a_{n+1}=a^{2}_{n}+\frac{1}{4}$ diverges to $+\infty$ This problem looks very interesting to me. It is a requirement to show that if

$$a_{n+1}=a_{n}^{2}+\dfrac{1}{4},\;n=1,2,3,\cdots\;\;\text{and}\;\;a_1=a,\;\;\text{then}$$
Case 1. $$\lim\limits_{n\to \infty}a_n=\dfrac{1}{2}\;\;\text{if}\;\;0<a\leq \dfrac{1}{2}.$$
Case 2. $$\lim\limits_{n\to \infty}a_n=\infty\;\;\text{if}\;\;a>\dfrac{1}{2}.$$
Proof of Case 1.
To prove 1, we need to show two things (i) monotonicity (ii) boundedness
(i) Monotonicity:
$$a_{n+1}-a_{n}=a_{n}^{2}+\dfrac{1}{4}-a_{n}=\left(a_{n}-\dfrac{1}{2}\right)^{2}\geq 0,$$
$$\implies a_{n}\leq a_{n+1}\;\;\forall\;\; n\geq 1.$$
Hence, $\{a_{n}\}$ is a monotone increasing sequence.
(ii) Boundedness:
Claim. $a_{n}\leq \dfrac{1}{2}\;\;\forall\;\; n\geq 1$
Proof of claim. The statement is true when $n=1$, since
$$a_{2}=a_{1}^{2}+\dfrac{1}{4}=a^{2}+\dfrac{1}{4}\leq \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}.$$
Assume it is true for $n=k$, that is
$$ a_{k+1}=a_{k}^{2}+\dfrac{1}{4}\;\;\text{for some}\;\; k\in N$$
Now, we prove it is true for $n=k+1$, i.e.,
$$ a_{k+2}=a_{k+1}^{2}+\dfrac{1}{4}=\left(a_{k}^{2}+\dfrac{1}{4}\right)^2+\dfrac{1}{4}\leq \dfrac{1}{2}.$$
Thus, by mathematical induction, it is true for all cases of $k\in N.$
Since it is monotone increasing and bounded above, it converges. The limit is easy to find. Let
$$\lim\limits_{n\to \infty}a_n=x,$$ then
$$x=x^{2}+\dfrac{1}{4}$$
$$\implies x=\dfrac{1}{2} \;\;(\text{twice}).$$
Thus, it is bounded by $1/2.$

Now, I have some issues with Case 2 but I'll give a useful definition.

Definition. A sequence $\{a_n\}$ is said to diverge to $+\infty$ if for any $M\in R,\;\exists\; n_M\in N\;$ such that
$$a_n> M\;\;\forall \;\;n\geq n_M.$$

My question is; how do I prove Case 2? If there are better ways of solving Case 1, I'll gladly welcome it. Corrections too, are welcome!
 A: Suppose that $a_n$ has a finite limit $l$, then we have : 
$$ \lim_{ n \rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_n \text{ hence if } a_n \text{ converges we have : } l = l^2 + \frac{1}{4}$$
This equation has the only solution $l = \frac{1}{2}$, hence if $a_0 > \frac{1}{2}$, because $a_n$ is clearly strictly increasing, we are done. (ie. $\lim_{n \rightarrow \infty} a_n = +\infty$).
A: You have already proved that if $(a_n)$ converges, then its limit is 1/2. You've also proved that $(a_n)$ is an increasing sequence. Now, since it is increasing and if $a_o =1/2$, then $1/2$ is a lower bound for the sequence and the sequence can never converge to it. This implies, that the sequence diverges to $+\infty$. It's positive because $(a_o)$ is positive and we have an increasing sequence.
A: Here's a hint on approaching the problem directly: consider the quantity $b_n=a_n-\frac12$.  Then you should be able to rewrite your recurrence in terms of $b_n$ and then find a simple inequality on $b_{n+1}$ in terms of $b_n$. (In particular, this will be another proof that $b_n$ is an increasing sequence!) Finally, using the fact that $b_1=a-\frac12\gt 0$, show that $b_n$ is unbounded; that is, for each $M$ you can find an $n_M$ such that $b_{n_M}\gt M$ (and thus $b_n\gt M$ for all $n\geq n_M$, since $b_n$ is an increasing sequence).
(Note that you won't be able to show uniform divergence; your $n_M$ will depend on the particular $a$ you're given as a starting condition. This is inevitable, because in fact the series isn't uniformly divergent — for any sufficiently large $M$ and any $n'$, there's some $a$ such that it will take you more than $n'$ steps for the sequence to get greater than $M$.)
