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If $(x_n)$ is any sequence of positive real numbers, then $$\lim \inf \dfrac{x_{n+1}}{x_n}\leq \lim \inf x_n^{1\over n} \leq \lim \sup x_n^{1\over n} \leq \lim \sup \dfrac{x_{n+1}}{x_n}. $$

I don't understand how am I suppose to use the hint nor what to do afterwards.

Hint: If $\lim \sup \dfrac{x_{n+1}}{x_n}-\infty$, right hand inequality is obvious. So suppose $\lim \sup \dfrac{x_{n+1}}{x_n}=M$. Then $\dfrac{x_{n+1}}{x_n}< M+\epsilon$ $\forall n\geq N$, i.e., $x_{n+k}\leq (M+\epsilon)^kx_n$ for all $k\geq 0$.

Yiorgos S. Smyrlis
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Username Unknown
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1 Answers1

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If $$ \limsup\frac{x_{n+1}}{x_n}=M<\infty, $$ then for every $\varepsilon>0$, there exists an $N\in\mathbb N$, such that $$ \frac{x_{n+1}}{x_n}\le M+\varepsilon, \quad n\ge N. $$ Thus $$ \frac{x_n}{x_N}=\frac{x_{N+1}}{x_N}\cdots\frac{x_{n}}{x_{n-1}}\le (M+\varepsilon)^{n-N}, $$ and hence $$ x_n\le \frac{x_N}{(M+\varepsilon)^N}\cdot (M+\varepsilon)^n\quad\Rightarrow\quad x_n^{1/n}\le \left(\frac{x_N}{(M+\varepsilon)^N}\right)^{1/n}\cdot (M+\varepsilon) \to M+\varepsilon. $$ Therefore $$ \limsup x_n^{1/n}\le M+\varepsilon,\quad \text{for all $\varepsilon>0$.} $$

Yiorgos S. Smyrlis
  • 83,933