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

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 ?

• What is the value of $u_0$ (or $u_1$)? Jul 22, 2020 at 16:06
• Any positive real number ( non nul) Jul 22, 2020 at 16:09

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 $$n\to +\infty$$.

• Yes indeed but what about the behaviour of $u_{n}$ and the series. Not yet clear. Jul 22, 2020 at 16:24
• @SeifS: I've expanded a bit the discussion. Do you have a proof of $u_n\sim\frac{C}{n}$ as $n\to +\infty$? Jul 22, 2020 at 16:37
• No don't have a proof but kind of verified numerically. But I struggle to find an equivalent for $u_{n}$. Mabe it is not even the right approach. Jul 22, 2020 at 16:40
• @SeifS: ok, I have proved that $u_n\sim\frac{\sqrt{6}}{n}$; the divergence of the series is now trivial. Jul 22, 2020 at 16:45
• I assume you have the equivalence form Gauss Integral but how did you get from $u_{n+1} \sim u_n - \frac{n}{6}u_{n}^{3}$ to $u_{n} \sim \frac{\sqrt(6)}{n}$ Jul 22, 2020 at 16:49

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 $$u_2\le u_1$$

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

• Yes indeed, $u_{n}$ is decreasing and we can show $0<u_{n}<1$ after a certain rank. Thus $u_{n}$ is convergent and we can show the limit is zero. But then I am unable to go further Jul 22, 2020 at 16:19
• @SeifS I've just added a little bit at the bottom. If you split this up into two series 1 is close to the the harmonic series and the other is close to a geometric series Jul 22, 2020 at 16:34
• I don't think that the representation through binomial coefficients is a good way to capture the magnitude of $u_n$, maybe it is better to exploit $$u_{n+1}\approx \int_{0}^{u_n} e^{-nt^2/2}\,dt \sim u_n-\frac{n}{6}u_n^3$$ Jul 22, 2020 at 16:40
• Good point Jack, you can show basically that $u_{n} - \frac{n}{6} u_{n}^{3}<u_{n+1}< u_{n} - \frac{n}{6} u_{n}^{3} +\frac{n(n-1)}{120} u_{n}^{5}$. But still not sure how this can bring us an equivalent of $u_{n}$ Jul 22, 2020 at 16:45
• @SeifS: the difference equation $u_{n+1}-u_n = -\frac{n}{6}u_n^3$ is analogous to the differential equation $f'(x)=-\frac{x}{6}f(x)^3$ which has simple, explicit solutions. $f(x)\sim\frac{\sqrt{6}}{x}$ allows to state $u_n\sim\frac{\sqrt{6}}{n}$. Jul 22, 2020 at 16:59