$\lim(x_{n+1}/x_n) = L \gt 1$. Show that $X$ is not a bounded sequence and hence is not convergent.
Suppose the sequence $X = x(n)$ is bounded. Then there exists a $M \in \mathbb{R}$ such that $|x_n| \leq M ~~\forall n \in \mathbb{N}$. Consider the set $$S := \{\textrm{all upper bounds of X}\}$$
$S$ is bounded below by $M$. By the completeness property, define $M := \inf(S)$. Define $L'$ such that $1 < L' < L$. Since $L' > 1$, then $M/L' < x_n \leq M ~~\forall n \in \mathbb{N}$ and $M < L' \cdot x_n$.
By the definition of limits, there exists a $K$ such that $n \geq K ~~\forall n \in \mathbb{N}$. Thus $x_{n+1}/x_n > L'~~ \forall n \geq K$ and $x_{n+1} > L' \cdot x_n$.
Combining the two inequalities together, we see that $$M < L' \cdot x_n < x_{n+1}$$
However this contradicts the fact that $|x_n| \leq M~~\forall n \in \mathbb{N}$ and that $M$ was defined as the greatest lower bound of the set $S$.
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I am not sure if I can simply define $M : = \inf(S)$ and that I am reaching the right contradiction that M was assumed to be the $\inf$ or if I am supposed to contradict that M was simply a lower bound of $S$.
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I also feel like my proof doesn't quite flow together as well as it should and was wondering how coherent my thoughts looked to other people.
Best Answer
Since $\lim_{n\to\infty} \frac{x_{n+1}}{x_n}=L>1$, we can choose a positive integer $N$ such that $\frac{x_{n+1}}{x_n}>\frac{1+L}2>1$ for $n\geqslant N$. Let $M>0$ and choose a positive integer $N'$ such that $N'>\frac{2M}{1+L}$. Then for $n\geqslant N+N'$ we have $$ \frac{x_{n+1}}{x_n} \geqslant \frac{x_{N+N'+1}}{x_{N+N'}} = \prod_{n=N}^{N+N'}\frac{x_{n+1}}{x_n} \geqslant N'\left(\frac{1+L}2\right) > M, $$ so that $\lim_{n\to\infty}x_n=+\infty$.