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Recently I encountered a strengthened version of this question. Suppose $x_1 > 0$ and $x_{n + 1} = \log(1 + x_n)$. Prove that $$\lim_{n \rightarrow \infty}\frac{n(na_n - 2)}{\log n} = \frac{2}{3}$$

I tried to follow the method for the original question. Define $b_n = 1/a_n$. It appears that $$b_{n + 1} = b_n + \frac{1}{2} - \frac{1}{12b_n} + O\left(\frac{1}{b_n^2}\right)$$ Therefore $$b_n = \frac{n}{2} - \frac{1}{12} \sum_{1 \leq k < n}\frac{1}{b_k} + O\left(\sum_{1 \leq k < n}\frac{1}{b_k^2}\right)$$ Now if we can prove that $$\sum_{1 \leq k < n}\frac{1}{b_k} = 2\log n + O(1)$$ $$\sum_{1 \leq k < n}\frac{1}{b_k^2} = O(1)$$ Then the limit follows easily, since $$\frac{n(na_n - 2)}{\log n} = \frac{n^2 - 2nb_n}{b_n\log n} = \frac{n/3 + O(n/\log n)}{b_n}$$ and we already know that $b_n/n \rightarrow 1/2$. The second asymptotic is already proved here. However I don't know if the first asymptotic is correct or not. We know that $b_n = n/2 + o(n)$. If we can show that the series $$\sum \frac{1}{n + o(n)} - \frac{1}{n}$$ converges, then we can conclude that $\sum 1/b_n = 2\ln n + O(1)$.

user2249675
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  • $\frac{1}{n+o(n)}-\frac{1}{n} = \frac{o(1)}{n+o(n)}$, but for example, take $\frac{1}{n\log(n)}$ and $\sum \frac{1}{n\log(n)}$ doesn't converge – Jakobian Dec 15 '18 at 16:28
  • Your claim does not hold in general, for instance when your $o(n)$ is very big, such as $o(n)=\frac{n}{\log(n)}$. – Aphelli Dec 15 '18 at 16:30
  • Maybe the condition could be weakened. We only need that $\frac{n - 2b_n}{\log n} \rightarrow 1/3$, hence $b_n = \frac{n}{2} - \frac{1}{6}\log n + o(\log n)$. – user2249675 Dec 15 '18 at 18:41

1 Answers1

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We have that

$$o(n)=n\cdot \omega (n) \quad \omega (n)\to 0$$

therefore

$$\frac{1}{n} - \frac{1}{n + o(n)}=\frac{\omega (n)}{n(1+\omega (n))}\sim \frac{\omega (n)}{n}$$

and the given series

  • converges for $\omega (n)=\frac1n$
  • diverges for $\omega (n)=\frac1{\log n}$
  • $\ldots$
user
  • 154,566
  • Small remark: be cautious with the alternating series, as convergence of $\sum n^{-1} \omega(n)$ does not imply convergence of $\sum n^{-1} (1+\omega(n))^{-1} \omega(n)$. – D. Thomine Dec 15 '18 at 21:06
  • @D.Thomine yes of course! I did a bad example with that...thanks! – user Dec 15 '18 at 21:09