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October 8th, 2017, 01:20 PM  #1 
Senior Member Joined: Dec 2006 Posts: 166 Thanks: 3  A limit has theoretical importance
If $a_1 = 1, a_{n+1}=\log(1+a_n),\;n\in\mathbb{N}^+$ Show that $\displaystyle{\lim_{n\to\infty}} \frac{a_n}{n a_n 2} = 0$ Last edited by greg1313; October 8th, 2017 at 04:19 PM. 
October 11th, 2017, 08:09 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 165 Thanks: 72 Math Focus: Dynamical systems, analytic function theory, numerics 
Here are a few hints which should get you close enough to finish it on your own. 1. The sequence $a_n$ is monotone decreasing. To see this, Taylor expand log and factor to see that the tail is negative. \[ a_{n+1} = \log(1+a_n) = a_n  \sum_{k = 2}^{\infty} \frac{1}{k!}a_n^k = a_n  \sum_{k = 1}^{\infty} a_n^{2k}(\frac{1}{(2k)!} + \frac{1} {(2k+1)!}a_n) < a_n \] 2. $a_n$ is nonnegative and thus the sequence is bounded below so it must converge to a limit. To see this, look at the same Taylor series from 1. 3. Next prove that $a_n \to 0$. To see this, notice that the mean value inequality requires the estimate \[ a_{n+1} = \log(1+a_n) \leq a_n \sup_{x \in (0,a_n)} \frac{1}{1+x}\] to hold for all $n$. Since the sequence is Cauchy, it follows that \[ \lim_{n \to \infty} \sup_{x \in (0,a_n)} \frac{1}{1+x} = 1 \] which requires $a_n \to 0$. The rest should be easy. 
October 12th, 2017, 09:53 AM  #3 
Senior Member Joined: Dec 2006 Posts: 166 Thanks: 3 
By Taylor, $0 = \log 1 < \log(1+x) = x\frac{x^2}{2!(1+\theta x)^2}< x\;\small(x > 0 < \theta < 1)$ So $0< a_{n+1} < a_n\;(\forall n),\;\{a_n\}$ is decreasing and bounded below hence convergent: $a_n\to A = \log(1+A)\ge 0\implies A=0$ i.e $a_n\searrow 0.$ By stolz and L'Hospital, ${\displaystyle\lim_{n\to\infty}} na_n={\displaystyle\lim_{n\to\infty}}\frac{n}{1/a_n}={\displaystyle\lim_{n\to\infty}}\frac{a_n\log (1+a_n)}{a_n\log(1+a_n)}=2$ Let $\tau_n = \frac{na_n2}{a_n} = n\frac{2}{a_n}$, then $a_n=\frac{2}{n\tau_n}\to 0,\;n\tau_n\to\infty$ Now $\tau_{n+1}\tau_n =1+\frac{2(\log(1+a_n)a_n)}{a_n\log(1+a_n)}= \frac{1}{6}a_n + O(a_n^2)$. Thus for $\small 0< \lambda < \frac{1}{6}$, $\tau_{n+1}\tau_n > \lambda a_n\;(n\ge m)$ with some $m$. And so $\tau_{n+1}> \tau_m+{\displaystyle\sum_{k=m}^n}\large\frac{2 \lambda}{k\tau_k}$ If $\{\tau_n\}_{n>m}$ were unbounded above, the RHS would be unbounded. a contradiction. So $\tau_{n+m}\nearrow \infty.$ Consequently ${\displaystyle\lim_{n\to\infty}}\frac{a_n}{na_n 2}={\displaystyle\lim_{n\to\infty}}\frac{1}{\tau_n }=0$ 
October 12th, 2017, 07:55 PM  #4 
Senior Member Joined: Dec 2006 Posts: 166 Thanks: 3 
This is only a small step of $\displaystyle{\lim_{n\to\infty}\frac{n(n a_n 2)}{\log n}=?}\;\;(a_ > 0,\; a_{n+1}=\log(1+a_n)$ The challenge is: Numeric experiment is very difficult for this kind of sequence 
October 13th, 2017, 05:37 AM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 165 Thanks: 72 Math Focus: Dynamical systems, analytic function theory, numerics 
1. Your claim that $a_n \to 0$ is flawed. You can't assume because 0 is a lower bound that the sequence necessary converges to 0 e.g. $a_n = \frac{1}{2} + \frac{1}{n}$. 2. In fact, my sketch of the proof that $a_n \to 0$ above is incorrect so I would have to think a bit more whether or not this is even true. However, after some thought it seems likely this isn't true and the quotient goes to 0 due to the help from the factor of $n$ in the denominator. 3. L'hospital's doesn't apply to sequences and Stolz doesn't apply to Cauchy sequences. Your analysis is heading in the right direction but looking directly at a quotient is problematic if indeed $a_n \to 0$. 4. Why is this hard to numerically simulate? This would certainly help determine if $a_n \to 0$ or not. 

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