Recursive sequence. Need help finding limit. This is my recursive sequence:
$a_1=\frac{1}{4};\space a_{n+1}=a_n^2+\frac{1}{4}$
for $n\ge 1$
In order to check if this converges I think I have to show that
1) The sequence is monotone increasing/decreasing
2) The sequences is bounded by some value
For 1) I am going to use the ratio test.
$\frac{a_{n+2}}{a_{n+1}}>1$ $\implies$ monotone increasing
$\frac{a_{n+2}}{a_{n+1}}<1$ $\implies$ monotone decreasing
$\frac{(a_{n+1})^2+\frac{1}{4}}{a_{n+1}}=a_{n+1}+\frac{1}{4}>0$ $\implies$monotone increasing
I am really not sure about this. How would I checkt/show it is bounded by some value?
 A: $a_n$ is increasing because
$$
\begin{align}
a_{n+1}-a_n
&=a_n^2-a_n+\tfrac14\\
&=(a_n-\tfrac12)^2\\
&\ge0
\end{align}
$$
If $a_n\le\frac12$, then
$$
\begin{align}
a_{n+1}
&=a_n^2+\tfrac14\\
&\le\frac12
\end{align}
$$
This implies that $a_n$ converges to some value less than or equal to $\frac12$. Let this value be $a$, then take the limit of the recursive equation
$$
\begin{align}
a
&=\lim_{n\to\infty}a_{n+1}\\
&=\lim_{n\to\infty}a_n^2+\tfrac14\\
&=a^2+\tfrac14
\end{align}
$$
which implies that $0=a^2-a+\frac14=(a-\frac12)^2$; that is, $a=\frac12$.
A: $a_{n+1}-a_n=\frac{(2a_n-1)^2}{4}> 0$. So this is a monotone increasing sequence. Now to see whether the sequence is bounded or not, observe that the limiting value should satisfy $a=a^2+1/4\implies a=1/2$. So, let the sequence be unbounded. Then $\exists N$ such that $a_{N-1}\le 1/2,\ a_N>1/2$. But $a_{N}>1/2\implies a_{N-1}>1/2$ which leads to a contradiction. Hence the sequence is bounded and converges to $1/2$.
A: It's easy to show that it's increasing by induction; $a_2=\frac1{16}+\frac14>\frac14>a_1$, and if $a_{n+1}>a_n$ for some $n\geqslant 1$, then 
$$a_{n+2}=a_{n+1}^2+\frac14>a_n^2+\frac14=a_{n+1} $$
(to be pedantic, you need to prove that $a_n>0$ for all $n$, but that is straightforward). Similarly, we can show that it is bounded above; clearly $a_1<\frac12$, and if $a_n<\frac12$  for some $n\geqslant 1$ then
$$a_{n+1}=a_n^2 + \frac14 < \left(\frac12\right)^2 + \frac14 = \frac12. $$
I'll leave it to you to find the limit.
