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 $2^m \le 3^n \le 2^{m+1} \Leftrightarrow 1 \le \frac{3^n}{2^m} \le 2$. 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 $[(1 + \frac{1}{10^{2014}})^k; (1 + \frac{1}{10^{2014}})^{k+1}]$, $k = 0, 1, \ldots, 10^{2014}$. 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. $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$.