<s>[INST] How to prove $0 < a_n < 1$ by induction I know $n \in \mathbb{N}$ and...
$$
a_n = \begin{cases}
0                         & \text{ if } n = 0 \\
a_{n-1}^{2} + \frac{1}{4} & \text{ if } n > 0
\end{cases}
$$


*

*Base Case:
$$a_1 = a^2_0 + \frac{1}{4}$$
$$a_1 = 0^2 + \frac{1}{4} = \frac{1}{4}$$
Thus, we have that $0 < a_1 < 1$. So our base case is ok.


*Inductive hypothesis:


Assume $n$ is arbitrary. Suppose 
$$0 < a_{n} < 1$$ 
$$0 < a_{n-1}^{2} + \frac{1}{4} < 1$$ 
is true, when $n > 1$.


*Inductive step:


Let's prove 
$$0 < a_{n+1} < 1$$
$$0 < a_{n}^{2} + \frac{1}{4} < 1$$
is also true when $n > 1$.
My guess is that we have to prove that $a^2_{n}$ has to be less than $\frac{3}{4}$, which otherwise would make $a_{n+1}$ equal or greater than $1$.
So we have $(a_{n-1}^{2} + \frac{1}{4})^2 < \frac{3}{4}$... I don't know if this is correct, and how to continue...
 [/INST] Ok I'll right it down so that it's clear to you :) 
I want to prove the property P: " 0 < $a_n$ < 1 "
I look at the property A: $0 < a_n < \frac{1}{2}$
A => P, that is : If A is true then P is true 
I'll prove A using induction (so technically I don't prove P by induction, but by implication).
$ a_o = 0  $ < $\frac{1}{2}$
If $ 0 < a_n < \frac{1}{2} $ , then :
$ a_{n+1} = a_n^2 + \frac{1}{4} $ > $a_n^2 $ > 0
And : $  a_{n+1} = a_n^2 + \frac{1}{4}  $ < $ (\frac{1}{2})^2 + \frac{1}{4} = \frac{1}{2} $
Hence you get : 0 < $a_{n+1}$ < $\frac{1}{2}$ :  the hypothesis holds for the rank n+1
So you have proven using induction that for every n positive integer you have :
0 < $a_n$ < $\frac{1}{2}$
But since : $\frac{1}{2}$ < 1 , you also have: 
0 < $a_n$ < $\frac{1}{2}$ < 1 ie 0 < $a_n$ < 1
</s>