Proof that if $\lim \frac{x_{n+1}}{x_n}=L>1$ then it diverges

Let $X=\{x_n\}$ be a sequence of strictly positive terms such that $\lim \frac{x_{n+1}}{x_n}=L>1$. I have to show that $X$ is unbounded and hence not convergent.

The simplest way to prove this is the following:

Lets assume that $L=\lim \frac{x_{n+1}}{x_n}>1$. Then for some $N$ such that $n \geq N$, $$\frac{x_{n+1}}{x_n}>1 \to x_{n+1} >x_n.$$ However if $x_{n+1} >x_n$ for all $n \geq N$ then $$\lim_{x \to \infty}a_n \neq 0$$ and hence by the divergent test $X=\{x_n\}$ is divergent.

Unfortunately I need to prove this by showing that is $X$ is unbounded. Can someone guide me?

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A sequence can very well be increasing and convergent. I don't know what you call the divergent test, but nothing can tell you that the sequence is divergent from what you have shown. – 1015 Jan 27 '13 at 21:36
Truesomeness :) – math101 Jan 27 '13 at 21:37

3 Answers

Write $L$ as $1+2k$ where $k>0$. Since $\frac{x_{n+1}}{x_n}\to L$, for some positive integer $N$, for all $n\geq N$, we have $1<1+k=L-k<\frac{x_{n+1}}{x_n}<L+k$.

It follows that $x_{N+m}>x_N(1+k)^m\geq x_N(1+km)$ for all positive integers $m$.

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By the definition of limit, there exists an $N$ such that if $n \ge N$ then $\frac{a_{n+1}}{a_n}\gt 1+\frac{L-1}{2}=1+d$.

Now you can argue that $x_{N+k}\gt x_N(1+d)^k$, since each time we increment $k$ by $1$, we multiply $x_k$ by at least $1+d$.

Then by the Binomial Theorem, or more simply by the Bernoulli inequality, $(1+d)^k \ge 1+kd$. That shows things blow up.

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Let $u>1$.

As $\frac{x_{n+1}}{x_n}<-u$ we see $x_{n}<-\frac{x_{n+1}}{u}$.

We choose $\epsilon<-\frac{x_{n+1}}{u}$.

As $\lim_{n \to \infty} x_n=0$, an N exists such that $|x_n|<\epsilon$.

Hence $-\epsilon<x_n<\epsilon$

Thus for $n>N$, $\frac{x_{n+1}}{x_n}<-u$

As the sequence $(\frac{x_{n+1}}{x_n})^\infty_{n=1}$ is unbounded, it is divergent

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