# Study convergence of series $\sum x_n$ with $x_{n+1}=\int_{0}^{x_n} \cos^{n}(t) dt.$

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

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

Let $$a= \sin x_1$$.

It follows that:

$$0, $$\;\;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 $$\;x_n=x_2=\int_0^{x_1} \cos t\;dt=\sin x_1= a>\frac{a}{2}=\frac{a}{n}$$.

Moreover:

$$0 and

$$0 for all $$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:

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

Therefore:

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

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 $$n\in \mathbb{N}-\{0\}.$$