series that diverges

Question

We consider the sequence $$(u_n)_{n\in\mathbb{N}}$$

defined by $0<u_0<1$ and $u_{n+1}=u_n-u_{n}^2$ for all $n\in\mathbb{N}.$

I want to prove that the serie with general term $\ln(\frac{u_{n+1}}{u_{n}})$ diverges.

Please help me to do so. This is what i showed: $u_n$ converges to 0 and $\frac{u_{n+1}}{u_{n}}$ converges to 1

$u_n\to 0$ and $\sum_n \ln(\frac{u_{n+1}}{u_{n}})$ is telescopic ... — Gabriel Romon, May 20 '19 at 21:25
There are several closely related MSE questions. For example MSE question 1558592 "Convergence rate of the sequence $a_{n+1}=a_n-a_n^2, a_0=1/2.$". Another is MSE question 2861768 "Calculate $lin_{n\to\infty} \frac{n(1-na_n)}{\log n}$" where $a_n$ is same as $u_n$ here. — Somos, May 20 '19 at 22:34

score 2 · Answer 1 · answered May 20 '19 at 21:25

2

Since $\ln\frac ab=\ln a-\ln b$, $$\sum_{n=0}^m\ln\frac{u_{n+1}}{u_n}=\ln u_{m+1}-\ln u_0\to -\infty$$

answered May 20 '19 at 21:25

can us prove that $u_n$ is equivalent to $1/n$? Thanks – user640960 May 20 '19 at 21:41
@user640960 I don't think you can, but why even bother about that? The above answers solves your question...! – DonAntonio May 20 '19 at 22:01
@user640960 It appears to be the case. You could try and call $y_n=1-nu_n$ for $n\ge1$, with the goal of proving that $y_n\to0$, and see where the fact that $y_{n+1}=\frac1{n^2}+\frac{n^2-n-2}{n^2}y_n+\frac{n+1}{n^2}y_n^2$ leads you. – May 20 '19 at 22:29
@Saucy O'Path? How is that ?can you please detail more your idea? thanks – user640960 May 20 '19 at 22:38
@user640960 You just substitute the original recursion in $1-(n+1)u_{n+1}$ and you obtain that recursion there. I haven't actually checked if $y_n\to 0$. – May 20 '19 at 22:48

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