Prove limit of a sequence If $s_n \neq 0$ and $L=\lim_{n\to\infty}\left|\frac{s_{n+1}}{s_n}\right|$, then by the definition of limit, we have $\forall n>N \implies \left|\frac{s_{n+1}}{s_n}-L\right| < \epsilon, \forall \epsilon > 0$. If $L < 1$, how do I start the proof?
The goal is to prove that if $L < 1$, then $\lim_{n\to\infty} s_n = 0$. The book says that I have to pick $a$, which is $L < a < 1$. Any hints?
 A: Let's see what's going on. Suppose that $L=0.77$. We are told that 
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L.$$
So after a while, the ratio $\left|\frac{a_{n+1}}{a_n}\right|$ is real close to $0.77$. In particular, after a while the ratio is always less than $0.9$.
What does after a while mean? Say it means for $n\ge 1000$. Let $|a_{1000}|=b$.
Since $\left|\frac{a_{1001}}{a_{1000}}\right|\lt 0.9$, we have
$$|a_{10001}|\lt 0.9b.$$
Since $\left|\frac{a_{1002}}{a_{1001}}\right|\lt 0.9$, we have
$$|a_{1002}|\lt (0.9)(0.9b)=(0.9)^2 b.$$
Similarly, $|a_{1003}|\lt (0.9)^3 b$, and in general 
$$|a_{1000+k}|\lt (0.9)^k b.$$
But $(0.9)^k\to 0$ as $k\to\infty$, so $|a_n|\to 0$ as $n\to\infty$. This implies that $a_n\to 0$ as $n\to\infty$.
A: Let $\varepsilon>0$. Since
$$
L=\lim_n\left|\frac{s_{n+1}}{s_n}\right|,
$$
there exists an $N \in \mathbb{N}$ such that 
$$
L-\varepsilon \le \left|\frac{s_{n+1}}{s_n}\right| \le L+\varepsilon \quad \forall n \ge N.
$$ 
If $0 \le L<1$, then there exists $\delta \in (L,1)$. Choosing $\varepsilon=\delta-L$ we get
$$
0 \le |s_n| \le |s_N|\delta^{n-N} \quad \forall n \ge N.
$$
Taking the limit we get $\lim_n|s_n|=0$, i.e. $\lim_ns_n=0$.
