Let $$F_n (x)=\int_0^x \cos^n (t)dt,\quad \forall n\in \mathbb N, \forall x\in \mathbb R$$ We define the sequence $ (x_n) $ by $$x_0\in]0,\pi[, \quad \forall n\in\mathbb N, x_{n+1}=F_n(x_n)$$ The objective is to study the convergence of this series $\sum x_n$.

First, we have $x_0=x_1$, $x_2>0$ and it's clear that $\forall n\geq 2,\quad 0\leq x_{n+1}\leq x_n \leq x_2 <\frac{\pi}{2}$, so the sequence $(x_n)$ converges and $\forall n\geq 2, x_n\leq \int_0^{\frac{\pi}{2}}\cos^n(t)dt\sim \sqrt{\frac{\pi}{2n}}$; thus $(x_n)$ converges to 0.

I have difficulties to continue studying the series

  • $\begingroup$ You can apply a comparison test in order to study the convergence of the series. $\endgroup$
    – Angelo
    Jul 3, 2020 at 9:52
  • $\begingroup$ @Angelo How do use this test $\endgroup$
    – Jane
    Jul 3, 2020 at 13:37

1 Answer 1


Let $a= \sin x_1$.

It follows that:

$0<a\le1$, $\;\;a\le x_1$.

I am going to prove by induction that:

$x_n\ge\frac{a}{n}$ for all $n\in \mathbb{N}-\{0\}.\;\;\;(*)$

For $n=1$, $(*)$ is true, indeed $\;x_n=x_1\ge a=\frac{a}{1}=\frac{a}{n}$.

For $n=2$, $(*)$ is true, indeed $\;x_n=x_2=\int_0^{x_1} \cos t\;dt=\sin x_1= a>\frac{a}{2}=\frac{a}{n}$.


$0<x_2=\sin x_1 \le 1$ and

$0<x_n\le x_2 \le 1$ for all $n\in \mathbb{N}, n\ge 2$.

Now I suppose that $x_n\ge\frac{a}{n}$ (where $n\ge2$) is true and prove that $x_{n+1}\ge\frac{a}{n+1}$.

$x_{n+1}=\int_0^{x_n} \cos^n t \; dt\ge\int_0^{\frac{a}{n}} \cos^n t \; dt\ge\frac{a}{n} \cos^n \left(\frac{a}{n}\right)$

Since $\;\cos x \ge 1 - \frac{x^2}{2}$ for all $x\in \mathbb{R}$, it follows that:

$\cos\left(\frac{a}{n}\right) \ge 1 - \frac{a^2}{2n^2}>0$.


$\cos^n\left(\frac{a}{n}\right) \ge \left(1 - \frac{a^2}{2n^2}\right)^n\ge 1-\frac{a^2}{2n}\ge 1-\frac{1}{2n}\ge 1-\frac{1}{n+1}$.

It follows that:

$x_{n+1}\ge\frac{a}{n} \cos^n \left(\frac{a}{n}\right)\ge\frac{a}{n}\left(1-\frac{1}{n+1}\right)=\frac{a}{n+1}$.

So by induction I have proved that:

$x_n\ge\frac{a}{n}$ for all $n\in \mathbb{N}-\{0\}.$


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