Nature of the series $\sum_{n=1}^\infty u_n$ where $u_{n+1} = \int_{0}^{u_{n}} \cos(x)^{n}dx$

Question

Can you find the nature of the series $\sum_{n=1}^\infty$ given by $u_{n+1} = \int_{0}^{u_{n}} \cos^n(x)dx$? You can show $u_{n}$ is convergent and the limit is 0. However it seems more difficult to find an equivalent to $u_{n}$ in order to study the series. The series seems divergent and $u_{n}$ evolves like $\frac{1}{n}$ but I am unable to find a proof. Any help ?

Answer 1

If $n$ is odd and $x\geq 0$ then $\int_{0}^{x}\cos^n(t)\,dt$ is non-negative and bounded by $1$, so up to index-shifting we may assume that $u_0\in(0,1]$. We have that $\{u_n\}_{n\geq 0}$ is decreasing and positive, so it is convergent. Decreasing since $$ u_{n+1} = \int_{0}^{u_n}\cos(t)^n\,dt < \int_{0}^{u_n}1\,dt = u_n.$$ Let us assume that $\lim_{n\to +\infty} u_n=L> 0$. Then for any $\varepsilon >0$ we have $$ L' = \int_{0}^{L''}\cos(t)^n\,dt $$ for any sufficiently large $n$, with $L'$ and $L''$ being at most $\varepsilon$-apart from $L$. Since $\cos(t)^n$ is pointwise convergent to zero on $(0,1)$, by the monotonic/dominated convergence theorem $L$ must be zero.

Since $u_n\to 0$,

$$ u_{n+1}\sim\int_{0}^{u_n}e^{-nt^2/2}\,dt \sim u_n-\frac{n}{6}u_n^3 $$ such that $u_n\sim\frac{\sqrt{6}}{n}$ from the solution of the separable ODE $$ f'(x) = -\frac{x}{6}f(x)^3 $$

and $\color{red}{\sum_{k=1}^{n}u_n\sim \sqrt{6}\log(n)}$ as $n\to +\infty$.

Answer 2

for an example, if we take $u_0=1$ $$u_1=\int_0^1dx=1$$ $$u_2=\int_0^1\cos(x)dx=\sin(1)\approx0.84$$ $$u_3=\int_0^{\sin(1)}\cos^2(x)dx=\frac{\sin(2\sin(1))+2\sin(1)}{4}\approx0.669$$ in fact, in general it is clear that $u_1=u_0$ and $u_2\le u_1$

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Another thought is that: $$u_{n+1}=\int_0^{u_n}\cos^n(x)dx=2^{-n}\int_0^{u_n}(e^{ix}+e^{-ix})^ndx=2^{-n}\sum_{r=0}^n{n\choose r}\int_0^{u_n}e^{-2irx}dx$$ $$u_{n+1}=2^{-(n+1)}\sum_{r=0}^n{n\choose r}\frac{1-e^{-2iru_n}}{ir}$$