$\lim_{n\to \infty }\frac{a_{n+1}}{a_{n}}< 1$, $a_{_{n}}> 0$- does $a_{_{n}}$ converge? I'd like your help with this:
The sequence $a_{_{n}}$ applies these condition:
$a_{_{n}}> 0$ for every $n \in \mathbb{N}$
$\lim_{n\to \infty }\frac{a_{n+1}}{a_{n}}< 1$.
I need to prove that $a_{n}$ is convergent,
and it's limit is 0.
I tried to work with the fact that $a_{_{n}}> 0$  and (not successfully) show that 
$a_{n}> a_{n+1}$, and than  to conclude what I need.
Thanks

Put $l:=\lim_{n\to+\infty}\frac{a_{n+1}}{a_n}$. We apply the definition of the limit with $\varepsilon :=\frac{1-l}2>0$. Hence we can find $n_0$ such that for $n\geq n_0$ $\frac{a_{n+1}}{a_n}\leq l+\frac{1-l}2 = \frac{l+1}2$. We get $0\leq a_{n+1}\leq a_n\frac{l+1}2$ hence $0\leq a_n\leq a_{n_0}\left(\frac{l+1}2\right)^{n-n_0}$ if $n\geq n_0$. Now you can conclude.
