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 $f(x)=\sin^4(t^2) \le 1.$

enter image description here

  • 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?


2 Answers 2


Since $\sin^4 x\le x^4$ and thus $$ 0\le a_{n+1}\le\int_0^{a_n}t^8\,dt=\frac19a_n^9 $$ one gets the estimate $$ 0\le a_{n}\le \sqrt[4]{3}\left(\frac{a_1}{\sqrt[4]3}\right)^{9^{n-1}}. $$ By the sandwich lemma, the limit is $0$.

  • $\begingroup$ Nice solution :) +1 $\endgroup$
    – RFZ
    Nov 5, 2015 at 9:36
  • $\begingroup$ elegant in its simplicity $\endgroup$ Nov 5, 2015 at 10:36

We have $$0\leq a_{2}=\int_{0}^{1}\sin^{4}\left(t^{2}\right)dt\leq\int_{0}^{1}dt\leq1 $$ then if we consider $$0\leq a_{n+1}=\int_{0}^{a_{n}}\sin^{4}\left(t^{2}\right)dt\leq a_{n}\leq1 $$ so the sequence is bounded, non negative and monotone decreasing.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .