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Let $\{x_n\}$ be a sequence of strictly positive real numbers. I want to show that

$$\lim \inf \frac{x_{n+1}}{x_n} \leq \lim \inf \sqrt[n]{x_n} .$$

I've tried and failed to assume the contrary to derive a contradiction. any hints?

Martin Garb
  • 157

1 Answers1

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Let $\liminf\dfrac{x_{n+1}}{x_{n}}=L$ and assume that $0<L<\infty$ and let $\epsilon\in(0,L)$, then for some $N>0$ we have $\dfrac{x_{n+1}}{x_{n}}>L$ for all $n\geq N$, so $\dfrac{x_{n+1}}{x_{N}}>(L-\epsilon)^{n-N+1}$, take $1/(n+1)$ both sides and $n\rightarrow\infty$, with the observation that $(L-\epsilon)^{(n-N+1)/(n+1)}\rightarrow L-\epsilon$, we are done.

user284331
  • 55,591