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Let $(a_n)_{n \geq 0} $ be a sequence with $ a_n \in \left(0, \frac{1}{3}\right) $, $ \lim_{n \to \infty} a_n = 0 $, and $ a_{n+1} = a_n (1 - a_n)(1 - 2a_n)(1 - 3a_n). $

Evaluate the limit:

$ \lim_{n \to \infty} \frac{n}{\ln n} \left( \frac{1}{6} - n a_n \right). $

My attempt was to apply Stolz-Cesaro but I kept encountering the case $0 \cdot \infty$. Then I tried using squeeze theorem by doing am-gm for $a_n (1 - a_n)(1 - 2a_n)(1 - 3a_n) $ but I was unsuccessful. Any help is appreciated.

math.enthusiast9
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1 Answers1

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This answer is heavily inspired by the first half of metamorphy's answer to $\{\log n-n(1-na_n)\}$ is convergent for $a_{n+1}=a_n(1-a_n)$ and $0<a_1<1$ (without using asymptotic analysis), which dealt with a similar, slightly simpler problem.

Define $b_n := 1/a_n$. One has $b_{n+1} \to +\infty$ and: $$\begin{split} b_{n+1}-b_n &= \frac{1 - (1-a_n)(1-2a_n)(1-3a_n)}{a_n(1-a_n)(1-2a_n)(1-3a_n)}\\ &= \frac{6a_n - 11a_n^2 + 6a_n^3}{a_n -6a_n^2 + 11a_n^3 - 6a_n^4}\\ &\xrightarrow[n \to \infty]{} \frac{6}{1} = 6\end{split}$$ Thus, by the SCT: $\displaystyle\frac{b_n}{n} \to 6$, and $na_n \to 1/6$.

Now, set $c_n := b_n - 6n$, so that $c_n = o(n)$, i.e. $c_n /n \to 0$, and: $$\begin{split} c_{n+1} - c_n &= b_{n+1} - b_n - 6 \\ &= \frac{6a_n - 11a_n^2 + 6a_n^3}{a_n -6a_n^2 + 11a_n^3 - 6a_n^4} - 6\\ &= \frac{25a_n^2 -60a_n^3 + 36a_n^4}{a_n -6a_n^2 + 11a_n^3 - 6a_n^4}\\ &\sim \frac{25a_n^2}{a_n} = 25a_n \sim \frac{25}{6n} \sim \frac{25}{6}(\log(n+1) - \log n)\end{split}$$

A transcription for those like OP who are unfamiliar with Landau notation and thus equivalence (and also a reminder for why this notation is useful):
We got that: $$\frac{c_{n+1} - c_n}{a_n} \to 25$$ thus, since $\frac{a_n}{1/n} \to \frac{1}{6}$ and $\frac{\log(n+1) - \log n}{1/n} = \frac{\log\left(1 + \frac{1}{n}\right)}{1/n} \to 1$, we have: $$\frac{c_{n+1} - c_n}{\log(n+1) - \log n} = \frac{c_{n+1} - c_n}{a_n} \cdot \frac{a_n}{1/n} \cdot \frac{1/n}{\log(n+1) - \log n} \xrightarrow[n \to \infty]{} 25 \cdot \frac{1}{6} \cdot 1 = \frac{25}{6}$$

By the SCT, this yields: $\displaystyle\frac{c_n}{\log n} \to 25/6$, therefore, since we have: $$a_n = \frac{1}{6n + c_n}$$ we get: $$\begin{split}\frac{n}{\log n}\left(\frac{1}{6} - na_n\right) &= \frac{n}{\log n}\left(\frac{1}{6} - \frac{1}{6 + \frac{c_n}{n}}\right)\\ &= \frac{n}{\log n} \frac{\frac{c_n}{n}}{6\left(6+\frac{c_n}{n}\right)}\\ &= \frac{c_n/\log n}{6(6 + c_n/n)}\\ &\xrightarrow[n \to \infty]{} \frac{25/6}{36} = \frac{25}{216}\end{split}$$

Bruno B
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  • Could you write the proof without little-o notation. I’m not very familiar with them – math.enthusiast9 Dec 02 '24 at 18:53
  • @math.enthusiast9 I've edited my answer to try to adhere to your request, hopefully it is clearer for you now. – Bruno B Dec 02 '24 at 19:26
  • It is clearer now. Thank you so much! – math.enthusiast9 Dec 02 '24 at 19:31
  • @math.enthusiast9 Glad to be of help! – Bruno B Dec 02 '24 at 19:32
  • @math.enthusiast9 $$c_{n+1} - c_n = \frac{25a_n^2 -60a_n^3 + 36a_n^4}{a_n -6a_n^2 + 11a_n^3 - 6a_n^4}$$ so that $$\begin{split}\frac{c_{n+1} - c_n}{a_n} = \frac{25a_n -60a_n^2 + 36a_n^3}{a_n -6a_n^2 + 11a_n^3 - 6a_n^4} = \cdot \frac{25 - 60 \cdot a_n + 36a_n^2}{1 -6a_n + 11a_n^2 - 6a_n^3} \to 25\end{split}$$ – Bruno B Dec 02 '24 at 20:39