a) Let $(a_n)$ be the sequence defined by $a_n=\ln (2n^2+1)-\ln (n^2+n+1)\,\,\forall n\geq 1.$ Prove that the set $\{n\in\mathbb{N}|\,\{a_n\}<\dfrac{1}{2}\}$ is a finite set; b) Let $(b_n)$ be the sequence defined by $a_n=\ln (2n^2+1)+\ln (n^2+n+1)\,\,\forall n\geq 1$. Prove that the set $\{n\in\mathbb{N}|\,\{b_n\}<\dfrac{1}{2016}\}$ is an infinite set.
Problem
Source: Vietnam Mathematical Olympiad 2016, Day 1, Problem 2
Tags: algebra, linear combination
talkon
06.01.2016 15:54
For part a), just note that $2>\frac{2n^2+1}{n^2+n+1}> \sqrt{e}$ for big enough n. So we must have $1> \ln 2 \geq \ln(\frac{2n^2+1}{n^2+n+1})> \frac{1}{2}$ for big enough n, so the set is finite
quangminhltv99
29.05.2016 04:55
Any solution for b)
ThE-dArK-lOrD
29.05.2016 19:04
Suppose to the contary that there exist $N\in \mathbb{Z}^+$ such that for all $n>N$, $\{b_n\} \geq \frac{1}{2016}$.
We have $2n^4<(2n^2+1)(n^2+n+1)<c\cdot 2n^4$ for large enough $n$ for all $c>1$.
Choosing $c=e^{\frac{1}{100000000}}$ gives $$\ln (2n^4)<b_n<\ln (2n^4)+\frac{1}{100000000}.$$Thus $\{ b_n\} =\{ \ln (2n^4)\}=\{ \ln (2) +2\ln (n^2)\}$ for all $n>M$ for some $M\in \mathbb{Z}^+$.
We have $ \{ 2\ln ((n^m)^2) \} =\{m\cdot 2 \ln (n^2)\}$ for all $m\in \mathbb{Z}^+$.
So $\{ \{ \ln (2) +2\ln (n^2)\} \mid n\in \mathbb{Z}^+,n>N,M \}$ is dense in $[0,1)$. Done.
Swag00
11.08.2016 16:10
Do the curly braces denote fractional part of x ?
khanhnx
28.06.2019 11:30
Here is my solution for part b Solution It's easy to see that: $(b_n)$ is increasing and $\lim b_n = + \infty$ We have: $\lim (b_n - b_{n - 1}) = \lim \ln \left(\dfrac{(2n^2 + 1)(n^2 + n + 1)}{(2n^2 - 4n + 3)(n^2 - n + 1)}\right) = 0$ Suppose there exist finite $n \in \mathbb{N}$ which satisfy $\{b_n\} < \dfrac{1}{2016}$ then there exist $n_0 \in \mathbb{N}$ which satisfies $\dfrac{1}{2016} \le \{b_n\} < 1, \forall n \ge n_0, n \in \mathbb{N}$ Since: $\lim (b_n - b_{n - 1}) = 0$ then there exist $n_1 \in \mathbb{N}$ which is big enough and satisfies: $b_n - b_{n - 1} < \dfrac{1}{2016}$ But: $\lim b_n = + \infty$ then there exist infinite $n > \max \{n_0; n_1\}$ which satisfies $[b_n] - [b_{n - 1}] = 1$ So: $[b_n] - [b_{n - 1}] + \{b_n\} - \{b_{n - 1}\} < \dfrac{1}{2016}$ or $\{b_{n - 1}\} - \{b_n\} > \dfrac{2015}{2016}$ Hence: $\{b_{n - 1}\} > \{b_n\} + \dfrac{2015}{2016} \ge \dfrac{1}{2016} + \dfrac{2015}{2016} = 1$, conflict with $\{b_{n - 1}\} < 1$ Therefore: there exist infinite $n \in \mathbb{N}$ which satisfy $\{b_n\} < \dfrac{1}{2016}$