Let $(c_n)^∞_{n=m}$ be a sequence of positive numbers. Then we have $\lim \inf_{n \to \infty} \frac{c_{n+1}}{c_{n}} \le \lim \inf c_{n}^{1/n}$
Let $L = \lim \inf_{n \to \infty} \frac{c_{n+1}}{c_{n}}$. If $L = -\infty$ then by definition $L \le \lim \inf c_{n}^{1/n}$.
Assume $L$ is a non-extended real number. Then for
there exists an $N \ge m$ such that $\frac{c_{n+1}}{c_{n}} \ge L-\epsilon$ for all $n \ge N$. Thus $c_{n+1} \ge c_{n}(L-\epsilon)$, $c_{n} \ge c_{N}(L-\epsilon)^{n-N}$ then denote $A = c_N(L-\epsilon)^{-N}, c_n =A(L-\epsilon)^n, c_n^{1/n} \ge A^{1/n}(L-\epsilon)$, $\lim_{n \to \infty} A^{1/n}(L-\epsilon) = L - \epsilon$. That is $\lim \inf_{n\to \infty}c_n^{1/n} \ge L - \epsilon$, since this is true for all $\epsilon$ than $\lim \inf_{n\to \infty}c_n^{1/n} \ge L$
Is this proof correct?
