# Prove that if $\lim_\limits{n\to \infty}\frac{x_{n+1}}{x_n}=L$, then $|L| \le 1$.

Let $x_n$ be a sequence that converges to $a$. ($a$ final limit)

prove that if $\lim_\limits{n\to \infty}\frac{x_{n+1}}{x_n}=L$, then $|L| \le 1$

any suggestions guys? I thought about proving by contradiction by suggesting that $L>1$ then by the convergence test it would mean that $x_n$ diverges to infinity. but the problem that the convergence test works only for positive sequences.

• If $a\neq 0$ then $L=1$. So the only case you are left with is $a=0$. – Thomas Andrews Dec 12 '15 at 20:54
• You need to prove that $|L|\le 1$. If $a\ne 0$, then we know that $L=1$. So, analyze the specific case $a=0$ and show that $|L|\le 1$ also. – Mark Viola Dec 12 '15 at 21:05
• If you are not trying to prove it, why are you asking this question? What? – Thomas Andrews Dec 12 '15 at 21:07
• sorry, Typo mistake, I meant this is what Im trying to prove, the special case $a=0$ .. but couldn't. does it mean that its equals zero because $x_{n+1}$ converges to zero faster? – F1sargyan Dec 12 '15 at 21:09
• If you knew the case $a\neq 0$, please put that in the question. You've caused me and Dr.MV to waste time telling you something you already knew. Help people help you. @F1sargyan – Thomas Andrews Dec 12 '15 at 21:11

HINT:

Assume that $\lim_{n\to \infty}x_n=0$ and $\lim_{n\to \infty}\frac{x_{n+1}}{x_n}=L$.

Show that for any given $\epsilon>0$, there exists a number $N(\epsilon)$ such that for $n>N$,

$$\min\left(|L-\epsilon|,|L+\epsilon|\right)^k\,|x_n| < |x_{n+k}| < \max\left(|L-\epsilon|,|L+\epsilon|\right)^k\,|x_n|$$

Note that as $k\to \infty$, $|x_{n+k}|\to 0$. Then conclude that $|L|\le 1$.

• Assuming $x_n$ is positive here, and that $L$ is non-negative. – Thomas Andrews Dec 12 '15 at 21:14
• @ThomasAndrews Yes. Good point. +1 for the catch. – Mark Viola Dec 12 '15 at 21:15
• Ok, I got what u did there, but what assures the expression $min\left(|L-\epsilon|,|L+\epsilon|\right)^k$ to be bounded for example ? @Dr.MV – F1sargyan Dec 12 '15 at 22:14
• If $|L|>1$, then there is a contradiction. – Mark Viola Dec 13 '15 at 2:50

Along the OP idea: assume that $\lim_{n\to \infty} \frac{ |x_{n+1}|}{|x_n|} > 1$. Then there exists $\epsilon > 0$ and $n_0$ so that $\frac{ |x_{n+1}|}{|x_n|} > 1+ \epsilon$ for all $n\ge n_0$, and so $|x_n| > (1+\epsilon)^{n-n_0}$ for all $n \ge n_0$, which implies $|x_n| \to \infty$.