Prove that if $\frac{x_{n+1}}{x_n}<1$ then $x_n$ converges to zero.

Question

Suppose that $x_n$ is a sequence of positive numbers and $\lim_{n\to\infty}{\frac{x_{n+1}}{x_n}}<1$. Prove that $x_n\to 0$.

I have an idea of how to prove this but first I need to know if there is the implicit assumption that the limit exists.

If I can use this assumption then I intend to use the definition of the limit to show that there is a number $r$ such that $\lim_{n\to\infty}{\frac{x_{n+1}}{x_n}}<r<1$. If I can choose epsilon to prove this statement then I will be able to write the proof.

Answer 1

Yes, when they say $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} < 1,$ it must mean that the limit exists. Otherwise how could it be less than one?

Answer 2

I think when you say "(...) the limit exists", you mean the limit of $x_n$, right?*

If so, then note that since $\lim \frac{x_{n+1}}{x_n}<1$, we have that there exists an $N$ such that if $n>N$, then $\frac{x_{n+1}}{x_n}<1$. Thus, $x_{n+1} <x_n$. It follows that $x_n$ is (eventually) a decreasing sequence, which is bounded below since the $x_n$'s are supposed to be positive. Thus, $x_n$ is indeed convergent.

*I believe this is what you meant because once you know that $\lim x_n=L$ for some $L$, then $\lim \frac{x_{n+1}}{x_n}=\frac{L}{L}=1$ if $L \neq 0$. Thus, $L$ must be $0$.