Prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = L < 1$ then $\lim_{n\to\infty} x_n = 0$ Let $(x_n)$ be a sequence with $x_n > 0$ for all $n \in \mathbb{N}$. I would like a hint on how to prove that if $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = L < 1$ then $\lim_{n\to\infty} x_n = 0$. 
Here is what i have tried. There exist a $N$ such that $|\frac{x_{n+1}}{x_n} - L| < \varepsilon$ for all $n \geq N$. From this i must conclude that for some $M$ we have $|x_n|<\varepsilon$ for $n \geq M$. Can anyone point me in the right direction? I suspect i need some smart trick. 
Thanks.

How about this reverse argument: since $\lim_{n\to \infty}x_{n+1}/x_n=L<1$, then, by the ratio test, the series
$$
\sum_{n=1}^\infty x_n < \infty
$$
That gives 
$$
\lim_{n\to \infty}x_n=0
$$
