It's given that: $$a_{n+1}=\int_{0}^{a_n} \sin^4(t^2) dt%$ $$ $$a_1=1$$
In order to prove that ${a_n}$ Converges I thought about showing that the sequence is monotonic (Increasing or Decreasing) and also that it is bounded (from above or below).
this is a graph of the function $f(x)=\sin^4(x)$ and we can deduce that $f(x)=\sin^4(t^2) \le 1.$
Monotonicity: I chose Induction method to show that the sequence is monotinic. by substituting $n=1$ :
$a_{2}=\int_{0}^{1} \sin^4(t^2) dt%$
($a_2$ must be positive because it describes the area below the function in the interval $[0,1]$.) $a_2$ describes an area that is surely less than $1$, and we can be sure by the graph.
$a_{3}=\int_{0}^{a_2} \sin^4(t^2) dt%$
its obvious that $a_3$'s area is less that $a_2$ and so on, this means that the sequence is monotonic decreasing. and it must be decreasing to zero.
- bounded series: also I want to show this by induction.
do you guys have any hints how to prove both things by induction?