Prove that $a_{n+1}=a_n^2+a_n^3$ $a_1=\frac{1}{2}$ converges (recursion sequence 
Prove that $a_{n+1}=a_n^2+a_n^3$ and $a_1=\frac{1}{2}$ converges (I just need this part and then I will continue the rest but I will put the full question anyway even if it is not needed)
and prove $b_{n+1}=b_n^2+b_n^3$ and $b_1=-1$ converges as well
let $c_n$ be the sequence $
c_n=
\begin{cases}
 a_n&\text{if}\, n _{even}\\
 b_n&\text{if}\, n_{odd}\\
\end{cases}
$
does $c_n$ converge?


We know that $a_1=\frac{1}{2}$ and $a_2=\frac{3}{8}$ , $a_3=\frac{99}{512}$ so the impression is that it is a decreasing sequence.
I did not prove that it is well defined because I think that it is trivial and there is not much to prove in this
I tried getting an impression of what the limit will be and did solved the following , assume that $a_n$ converges then $L=L^2+L^3$ and after solving I got $L_1=0$ ,$L_2=\frac{1+ \sqrt5}{2}$ , $L_3=\frac{1-\sqrt5}{2}$
I am not sure if this is right but from this we can get $0 \leq a_n \leq \frac{1+ \sqrt5}{2}$
I believe that here I should prove that the sequence is bounded and monotone because we know that every sequence that is monotone and bounded converges but how can I prove that it is decreasing and bounded ?
I tried induction and I did not know how to continue
Induction basis for $n=1$ - $a_1=\frac{1}{2}$ and $a_2=\frac{3}{8}$ as mentioned above
Induction proposition - for $n=k \geq 2$ - to show that the sequence is decreasing $a_{k-1} \geq a_k$
Induction step prove for $n=k+1$ - $a_k \geq a_{k+1}$
so we need to prove $a_k \geq a_{k+1}$ and from what we assume in the induction $a_{k-1} \geq a_k$  so $a_{k-1} \geq a_k \geq a_{k+1}$ $\to$ $a_{k-1} \geq a_{k+1}=a_k^2 +a_k^3$ but how do I continue from here (if it is right)?
Thanks for any help and tips hopefully the translations are understandable
Edit - I now believe that what I did with the $"L"$ is not correct because if a sequence is decreasing then it is bounded from above by $a_1$ and not the $L$ that I found
 A: Let $f : \left[0, \dfrac{1}{2} \right] \rightarrow \mathbb{R}$ defined by $f(x)=x^2+x^3$. Then $f$ is increasing (as the sum of two increasing functions), so
$$f \left(\left[0, \dfrac{1}{2} \right] \right) = \left[f(0), f\left(\dfrac{1}{2}\right) \right] = \left[0, \dfrac{3}{8} \right] \subset \left[0, \dfrac{1}{2} \right]$$
So the sequence $(a_n)$ is well-defined and takes its values in the interval $\left[0, \dfrac{1}{2} \right]$.
Moreover, for every $x \in \left[0, \dfrac{1}{2} \right]$, one has
$$f(x)-x = x^2 + x^3-x = x\left(x + \dfrac{1+\sqrt{5}}{2} \right)\left(x + \dfrac{1-\sqrt{5}}{2} \right) \leq 0$$
so $f(x) \leq x$, so for every $n$, $f(a_n) \leq a_n$, so $a_{n+1} \leq a_n$.
So the sequence $(a_n)$ is decreasing, and because it is bounded below by $0$, then it converges (and it converges to $0$ since $0$ is the only fixed point of $f$ in the given interval).
A: If all you need to prove is convergence for $a_1=1/2$, show that the sequence is monotonic and bounded.

*

*With $a_1$ positive, every term is forced positive, giving a lower bound of $0$.


*Render $a_{n+1}/a_n=a_n(1+a_n)$, which is $<1$ for $0\le a_n \le 1/2$.
