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
$$(n-1)-\left(\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\cdots+\dfrac{a_{n-1}}{a_{n}}\right)>1$$
 A: 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 $n > k$.
A: Let $x_k=\frac{a_{k+1}-a_k}{a_{k+1}}>0$. There are two cases:


*

*If the sequence $(x_k)_k$ does not converge to $0$ then clearly $\sum_{k=1}^\infty x_k=+\infty$.

*If $~\lim_{k\to\infty}x_k=0$, then $$x_k\sim_{+\infty} -\ln(1-x_k)=\ln(a_{k+1}/a_k)$$ and consequently, $\sum_kx_k$ and $\sum_k\ln(a_{k+1}/a_k)$ have the same nature, but the later is divergent since $(a_k)_k$ is increasing and unbounded. So in this case we have also $\sum_{k=1}^\infty x_k=+\infty$. 


It follows that there is some $k>0$ such that for $n>k$ we have
$\sum_{k=1}^\infty x_k>2014$ and this is equivalent to the proposed inequality.
A: An "elementary" proof:
The statement to be proved can be written as 
$${a_2 - a_1 \over a_2} + {a_3 - a_2 \over a_3} + ... + {a_n - a_{n-1} \over a_n} > 2014$$
This is to hold for $n$ sufficiently large. Since the terms are positive, it suffices to show that
$$\sum_{n = 1}^{\infty} {a_{n+1} - a_n \over a_n} = \infty$$
Consider the sum of the terms of the above sequence with denominators ranging from $a_k$ to $a_l$ for some $l > k$. Since the $a_n$ increase, this is greater than
$${a_{k+1} - a_{k} \over a_{l}} + {a_{k+2} - a_{k+1} \over a_{l}} + ... + {a_{l} - a_{l- 1} \over a_{l}}$$
$$= {a_{l} - a_k \over a_{l}}$$
Since the $a_n$ increase to $\infty$, we can find a sequence $n_1 < n_2 < ...$ such that $a_{n_{i+1}} > 2a_{n_i}$ for each $i$. Then by above the terms of the sequence from $k = n_i$ to $l = n_{i+1}$ add up to at least 
$${a_l - {1 \over 2}a_l \over a_l}$$
$$= { 1\over 2}$$
Since there are infinitely many such blocks of terms, which can be made disjoint by just looking at every other block, the sum is infinite.
