Sequence convergence to zero Suppose that {$x_n$} is a sequence of positive numbers and 
$\hspace{70mm}$ lim ($\frac {x_{n+1}}{x_n}$) = $L$ 
Show that  if $ L < 1$, then the lim $x_n = 0$
We know that if $L < 1$ then ${x_{n+1}}$ must be strictly less than ${x_n}$ and if ${x_{n+1}} < {x_{n}}$ for every $n$, then the sequence is monotone decreasing. Since the sequence is of all positive numbers, it is bounded below by zero, it's infimum. Is this sufficient? 
 A: Not sufficient. A sequence with $a_n=\frac{1}{2} + \frac{1}{n}$ is decreasing, bounded below by zero, monotone, and contains all positive numbers. Its limit is not zero.
Of course, its infimum is also not zero, but you haven't shown that $0$ is the infimum for $x_n$ either.
Hint: If $\frac{x_{n+1}}{x_n}$ would be $L$, then $x_{n+1}=L^nx_0$ which has a limit of $0$. Of course, you don't have $\frac{x_{n+1}}{x_n}=L$, but can you find a $L'<1$ such that you have $\frac{x_{n+1}}{x_n}<L'$ at least for $n>N$ for some $N$?
A: This was my first approach, but ultimately incorrect:
Consider the following:
$$ 0 < x_{n+1} < L x_n < L^2 x_{n-1} < ... < L^{n+1} x_0 $$
So, we ultimately have
$$ 0 < x_{n+1} < L^{n+1} x_0 $$
Now use squeeze theorem. :-)
(This is wrong because the ratio is only true in the limit; without carefully considering some $\epsilon$'s, this is no good.)
Try this approach instead
Try instead a contradiction -- suppose $L<1$ but $x_n \to c \neq 0$. Then, $x_{n+1}/x_n\to1$. Since $L\neq1$ (and we know we can't go off to infinity due to boundedness), we're done here.
** [Edit] That's wrong too, I think. **
