Assume all $s_n \ne 0$ and that the limit $L = \lim\left|\frac{s_{n+1}} {s_n}\right|$ exists. How to show that if $L < 1$, then $\lim s_n = 0$?
There is one suggested solution, but I don't quite get it. It goes as the following:
Let $a$ be such that $L<a<1$. Let $\epsilon=a-L$, then there exists a $N$ such that $n>N\implies \left|\left|\frac{s_{n+1}}{s_n}\right|-L\right|<\epsilon=a-L$, then $|\frac{s_{n+1}} {s_n}|<a\implies |s_{n+1}|<a|s_n|$. This also implies that $|s_{N+1}|<a|s_N|$ and $|s_{N+2}|<a^2|s_{N}|$ and so on to $|s_n|<a^{n-N}|s_N|$ . Since $|a|<1$, $a^{n-N}\to0$. Thus $\lim s_n=0$
$1.$ Why are we allowed to arbitrarily set $\epsilon$? I thought we have to account for any $\epsilon.$
$2.$ if $a^{n-N}\to0$, then $|s_n|<0$, how does $\lim s_n$ get to be $0$? it seems it's negative.
$3.$ we have $2L-a<L-\epsilon<\left|\frac{s_{n+1}}{s_n}\right|<L+\epsilon<a$, but we only account for the right side but not the left side.
$(1)$ We pick $\varepsilon>0$ (allowed to be arbitrarily small) to invoke $L=\lim\left|\frac{s_{n+1}}{s_n}\right|$, which is given to us.
$(2)$ This is the squeeze/sandwich theorem from calculus. Since $0\leq|s_n|< a^{n-N}|s_N|$ and $\lim_na^{n-N}|s_N|=0$, we have $\lim_n|s_n|=0$.
$(3)$ You don't need $2L-a<L-\epsilon<\left|\frac{s_{n+1}}{s_n}\right|$ to obtain the result, so why mention it?