Convergence of sequence $a_{n+1} = \int_0^{a_n}[1+\frac{\cos^{2n+1} t}{4}]dt$

Question

How to show that sequence $$a_{n+1} = \int_0^{a_n}\left(1+\frac{\cos^{2n+1} t}{4}\right)\,dt$$ with $a_0 \in (0, 2\pi)$ is convergent?

Answer 1

Note that $$\tag1a_{n+1}-a_n = \int_0^{a_n}\frac{\cos^{2n+1}t}4\,\mathrm dt$$ so that (using $|\cos|\le 1$ and $\cos(\pi- x)=-\cos x $ etc., $$\begin{align}0\le a_{n+1}-a_n&\le \frac14a_n,& 0\le a_n\le\frac\pi2\\ 0\le a_{n+1}-a_n&\le \frac14(\pi-a_n),& \frac\pi2\le a_n\le\pi\\ 0\ge a_{n+1}-a_n&\ge - \frac14(a-\pi),& \pi\le a_n\le\frac32\pi\\ 0\ge a_{n+1}-a_n&\ge -\frac14(2\pi-a_n),& \frac32\pi\le a_n\le2\pi.\\ \end{align}$$

In particular $$ a_n\in[0,\pi]\implies a_n\le a_{n+1}\in[0,\pi]$$ $$ a_n\in[\pi,2\pi]\implies a_n\ge a_{n+1}\in[\pi,2\pi]$$ At any rate, $\{a_n\}_n$ is bounded and monotonic, hence convergent.

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Remark: We see from the above that the slightly "wilder" recursion without the denominator $4$ leads to the same convergence result. In fact, even $$ a_{n+1}=\int_0^{a_n}\left(1+q\cos^{2n+1}t\right)\,\mathrm dt$$ with $0<q<2$ and $a_0\in(0,2\pi)$ gives us convergence (though the limit may be $>2\pi$).