show the condition $\lim_{n\rightarrow \infty} \frac{s_{n+1}}{s_n} <1$ implies that $s_n \rightarrow 0$ Let $\{s_n\}$ be a sequence of positive numbers. Show that the condition
$\displaystyle \lim_{n\rightarrow \infty} \frac{s_{n+1}}{s_n} < 1$
implies that $s_n \rightarrow 0$.
I cannot think of how to start. Can someone help? Thank you.
 A: Hint:
(First assume that the sequence converges.)
Assume for contradiction that $\displaystyle \lim_{n\rightarrow \infty} s_n =a$ where $a>0$. 
What can you say about  $\displaystyle \lim_{n\rightarrow \infty} s_{n+1} $?
What can you then say about  $\displaystyle \lim_{n\rightarrow \infty} \frac{s_{n+1}}{s_n} $?
(Then see what happens if the sequence diverges. Is it possible?)
A: *

*Since $\lim \frac{s_{n+1}}{s_n}=L <1$ it follows that for $n$ great enough $s_{n+1}/s_n <(L+1)/2<1$. This gives you the monotonicity of $(s_n)$ for $n$ great enough.

*$(s_n)$ is decreasing and bounded by $0$, therefore...

*If $s_n \to a>0$ then the limit tells you $1<1$, therefore...
A: Consider the series: s(1) + s(2) + s(3) +...+ s(n) +..... The condition that lim s(n+1)/s(n) < 1 implies that this series converges by ratio test. This means the n-th term s(n) must have
limit = 0.
A: Even more: if for some constant $0<c<1$:
$$\frac{s_{n+1}}{s_n}\le c$$
for $n$ large enough then
$$0\le s_n\le Mc^n$$
and
$$\lim_{n\to\infty}s_n=0.$$
