# Prove that the following series is divergent

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.

• but the k+1 in curly brackets { } to make them subscript properly $a_{k+1}$ – Doug M Jun 6 '18 at 21:36

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 $n>m$.

• I agree that OP's observation shows that partial sums of $(x_k)$ is not Cauchy, but we need the full power of the lower bound $1-\frac{a_m}{a_n}$ rather than $1-\frac{a_m}{a_{m+1}}$. (Consider $a_n = n$, for instance.) – Sangchul Lee Jun 6 '18 at 21:42
• ${1-{m\over{m+1}}}>0$. – Tsemo Aristide Jun 6 '18 at 21:45
• Ah you are right. I must be half-awake. (+1) – Sangchul Lee Jun 6 '18 at 21:45
• Thank you so much sir – Normal Jun 6 '18 at 21:54
• +1: Nice, simple answer. – copper.hat Jun 7 '18 at 17:25

It can be seen as a particular case of this result :

If the positive series $\sum a_n$ diverges and $s_n=\sum\limits_{k\leqslant n}a_k$ then $\sum \frac{a_n}{s_n}$ diverges as well

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 $n\to \infty.$