Prove that $\exists k>0$ such that$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}
Consider a positive sequence $\{a_{n}\}$ such that  $a_{n+1}>a_{n}$, and $\{a_n\}$ is unbounded.
Show that there exits a positive integer $k$ such that, when $n>k$
  $$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}<n-2014$$
This problem is from china test 2014.
I only prove follow: there exsit $k$,such $n>k$,have
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}<(n-1)-1$$
because
\begin{align*}
&(n-1)-\left(\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}\right)\\
&=\dfrac{a_{2}-a_{1}}{a_{2}}+\dfrac{a_{3}-a_{2}}+\cdots+\dfrac{a_{n}-a_{n-1}}{a_{n}}\\
&>\dfrac{a_{2}-a_{1}}{a_{2}}+\dfrac{a_{3}-a_{2}}{a_{n}}+\cdots+\dfrac{a_{n}-a_{n-1}}{a_{n}}
&=\dfrac{a_{2}-a_{1}}{a_{2}}+\dfrac{a_{n}-a_{2}}{a_{n}}\\
&=1+a_{2}\left(\dfrac{a_{2}-a_{1}}{a^2_{2}}-\dfrac{1}{a_{n}}\right)
\end{align*}
since $a_{n}$ is unbounded,so we choose $\{a_{n}\}$ such
$$a_{n}>\dfrac{a^2_{2}}{a_{2}-a_{1}}$$
so
$$(n-1)-\left(\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}\right)>1$$

Let $(1-\epsilon_n) = a_{n} / a_{n+1}$,  Then $a_n = a_1 / \prod_{i=1}^{n-1} (1-\epsilon_i)$.  Since $a_n$ is unbounded, it follows that $\prod_{i=1}^\infty (1-\epsilon_i) = 0$.  It is well known that this product equals zero if and only if $\sum_{i=1}^\infty \epsilon_i$ diverges.
But $\frac{a_1}{a_2} + \cdots \frac{a_{n-1}}{a_n} = n - \sum_{i=1}^{n-1} \epsilon_i$, and since $\sum_{i=1}^\infty \epsilon_i$ diverges, for some integer $k$ we have that  $\sum_{i=1}^{n-1} \epsilon_i > 2014$ for $n > k$.
