Mongolian Mathematical Olympiad

From the Euler number's definition $(1+\frac{1}{10^{2014}}{)}^{10^{2014}}>2$.

For any nonnegative whole number $n$, there exists nonnegative $m$ such that $2^m\le 3^n\le 2^{m+1}$. Different $n_1,n_2$ correspond to different $m_1,m_2$. Note that $2^m\le 3^n\le 2^{m+1}\iff 1\le \frac{3^n}{2^m}\le 2$. There exist infinitely many different numbers $\frac{3^n}{2^m}$ that lie in the segment $[1;2]$.

Divide the segment $[1;2]$ into small segments $[(1+\frac{1}{10^{2014}}{)}^k;(1+\frac{1}{10^{2014}}{)}^{k+1}]$, $k=0,1,\dots ,10^{2014}$. Since there are infinitely many numbers of the form $\frac{3^n}{2^m}$, by the pigeonhole principle there is a segment inside in which lie two different numbers $\frac{3^{n_1}}{2^{m_1}},\frac{3^{n_2}}{2^{m_2}}$. Let us denote this segment by $s$. In other words, $${\left(1+\frac{1}{10^{2014}}\right)}^s\le \frac{3^{n_1}}{2^{m_1}}<\frac{3^{n_2}}{2^{m_2}}\le {\left(1+\frac{1}{10^{2014}}\right)}^{s+1}.$$ From this it follows that $1<\frac{3^{n_2}}{2^{m_2}}:\frac{3^{n_1}}{2^{m_1}}\le 1+\frac{1}{10^{2014}}$ and $1<\frac{3^{n_2}{2}^{m_1}}{2^{m_2}{3}^{n_1}}<1+\frac{1}{10^{2014}}$.

Setting $a_{n+q}=3^{n_2}{2}^{m_1}$, $a_n=3^{n_1}{2}^{m_2}$ we get $1+\frac{1}{10^{2014}}>\frac{a_{n+q}}{a_n}\ge \frac{a_{n+1}}{a_n}$.

Continuing in this manner from the fraction $\frac{a_{n+1}}{a_n}$ it is possible to construct a new fraction less than $1+\frac{1}{10^k}$ for any natural $k$.

Note. ${\lim }_{n\to \infty }\frac{a_{n+1}}{a_n}=1$.