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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 ?

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

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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 $n\to +\infty$.

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  • $\begingroup$ Yes indeed but what about the behaviour of $u_{n}$ and the series. Not yet clear. $\endgroup$
    – Seif S
    Jul 22, 2020 at 16:24
  • $\begingroup$ @SeifS: I've expanded a bit the discussion. Do you have a proof of $u_n\sim\frac{C}{n}$ as $n\to +\infty$? $\endgroup$ Jul 22, 2020 at 16:37
  • $\begingroup$ 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. $\endgroup$
    – Seif S
    Jul 22, 2020 at 16:40
  • $\begingroup$ @SeifS: ok, I have proved that $u_n\sim\frac{\sqrt{6}}{n}$; the divergence of the series is now trivial. $\endgroup$ Jul 22, 2020 at 16:45
  • $\begingroup$ 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 $u_{n} \sim \frac{\sqrt(6)}{n}$ $\endgroup$
    – Seif S
    Jul 22, 2020 at 16:49
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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: $$u_{n+1}=\int_0^{u_n}\cos^n(x)dx=2^{-n}\int_0^{u_n}(e^{ix}+e^{-ix})^ndx=2^{-n}\sum_{r=0}^n{n\choose r}\int_0^{u_n}e^{-2irx}dx$$ $$u_{n+1}=2^{-(n+1)}\sum_{r=0}^n{n\choose r}\frac{1-e^{-2iru_n}}{ir}$$

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  • $\begingroup$ 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 $\endgroup$
    – Seif S
    Jul 22, 2020 at 16:19
  • $\begingroup$ @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 $\endgroup$
    – Henry Lee
    Jul 22, 2020 at 16:34
  • $\begingroup$ 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 $$u_{n+1}\approx \int_{0}^{u_n} e^{-nt^2/2}\,dt \sim u_n-\frac{n}{6}u_n^3$$ $\endgroup$ Jul 22, 2020 at 16:40
  • $\begingroup$ 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 $u_{n}$ $\endgroup$
    – Seif S
    Jul 22, 2020 at 16:45
  • $\begingroup$ @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 $u_n\sim\frac{\sqrt{6}}{n}$. $\endgroup$ Jul 22, 2020 at 16:59

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