Let $\{a_k\}_k $ be an unbounded, strictly increasing sequence of positive real numbers and $x_k = \frac{a_{k+1}- a_k}{a_{k+1}}$ then show that $\sum_{k=1}^\infty {x_k} $ is divergent.
I proved that for all $n \geq m$, $\sum_{k=m}^{n} x_k \geq 1-\frac{a_m}{a_n}$. How I apply this to show divergent of that series, I can't understand.
The result that you have proven implies that $\sum x_k$ is not a Cauchy sequence since $1-{a_m\over a_n}\geq 1-{a_m\over{a_{m+1}}}>0$ for $n>m$.
It can be seen as a particular case of this result :
Let's take $b_n=a_{k+1}-a_k$ then $s_n=\sum\limits_{k=0}^n b_k=a_{n+1}-a_0=a_{n+1}\to+\infty\quad$ unbounded hypothesis
(without loss of generality let assume $a_0=0$)
Then $\sum x_n=\sum \frac{b_n}{s_n}=\sum \frac{a_{n+1}-a_n}{a_{n+1}}$ diverges as well.
Two useful facts about series, both easy to prove:
(1). If each $x_k\geq 0$ then $\sum_{k\in \Bbb N}x_k$ converges iff $\prod_{k\in \Bbb N}(1+x_k)$ converges.
(2). If each $x_k\in [0,1)$ then $\sum_{k\in \Bbb N}x_k$ converges iff $0\ne \prod_{k\in \Bbb N}(1-x_k).$
Use (2) for this Q. We have $\prod_{k=1}^n(1-x_k)=a_1/a_{n+1}\to 0$ as $n\to \infty.$
