Help proving exercise on sequences in Bartle's Elements Self learning Analysis and found the following exercise in Bartle's Elements of Real Analysis:

Let $X = (x_n)$ be a sequence of strictly positive real numbers such
  that $\lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1.$Show that
  for some $r$ with $0\lt r \lt 1$ and some $C \gt 0 $ we have $ \ 0 \lt
 x_n \lt Cr^n$ for all sufficiently large $n \in \Bbb N$. Use this to
  show that $\lim(x_n) = 0.$

Now I haven't proven any significant results concerning sequences and hence am required to solve this problem with basically nothing but the definition of a limit. 
I am not getting too far with this because I cannot see a way to get there with a direct implication and I cannot for the life of me negate "for all sufficiently large $n$" in the context of the sentence above. 
Would be extremely grateful if someone could help me out.
What I've done so far:
We know that $l = \lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1$. If $l \lt 0$ we can find a neighbourhood of $l$ entirely contained in $(-\infty, 0)$ which would contain no elements of the sequence $  \left({\frac {x_{n + 1}} {x_n}} \right) $ all of whose elements are positive leading to a contradiction. This implies $l \ge 0.$
We can pick $ 0 \lt  \epsilon \lt 1 - l$. There is $m \in \Bbb N$ such that 
$$n \ge m \implies \left|{\frac {x_{n + 1}} {x_n} - l}\right| \lt \epsilon \implies \frac {x_{n + 1}} {x_n} \lt \epsilon + l \lt 1 \implies x_{n + 1} \lt x_n$$ 

Also can someone tell me if $\lim (x_n) = 0$ necessarily? Doesn't
  $\left({2 + \frac 1 n}\right) $ satisfy the given conditions??


You have successfully unpacked the definition of the limit into the form that will become useful. You have already shown that there exists $m\in\mathbb N$ such that for all $n\geq m$ you have
$$\frac{x_{n+1}}{x_n}<l+\epsilon=:r<1. \tag1$$
All you need now is to put some recursion to work. This is the core of the criterion: if the limit of ratios of successive elements is smaller than one, this means that each element of the sequence is smaller than the previous one by some fixed percentage. If you go multiple steps out, this percentage will be compounded, which is what gets you the bound.
To make that precise, note that multiple applications of the bound (1) will give you


*

*$x_{m+1}<x_m\,r,$

*$x_{m+2}<x_{m+1}\,r<x_m\,r^2,$

*$x_{m+3}<x_{m+2}\,r<x_m\,r^3,$


and so on, so that a trivial induction step will give you 
$$x_{m+k}<x_m\,r^k.$$
This is essentially the bound you were after, if you set $C=x_m/r^m$.
