Proof of limit inequality

Question

Prove that for any sequence $\{x_n\}$ of positive real numbers $$\lim\text{sup}\sqrt[n]{x_n}\leq \lim\text{sup}\frac{x_{n+1}}{x_n}.$$

My attempt:

Let $A = \lim\text{sup}\frac{x_{n+1}}{x_n}$. Suppose $A<\infty$ and choose $\epsilon >0$, then $\exists$ an integer $N$ so that $N\le n \implies \frac{x_{n+1}}{x_n}\le \epsilon$. But I do not know how to proceed in finishing the proof?

No, that's wrong. The conclusion could be $\dfrac{x_{n+1}}{x_n} \le A + \epsilon$. — Robert Israel, Mar 05 '13 at 07:56

score 2 · Accepted Answer · answered Mar 05 '13 at 07:57

2

Hint: if $x_{n+1}/x_n \le B$ for $n \ge N$, then $x_n \le x_N B^{n-N}$ for $n \ge N$. What does that tell you about $x_n^{1/n}$?

answered Mar 05 '13 at 07:57

Robert Israel

448,999

Since $x_n \le x_N B^{n-N}$ then $x_n^{1/n}\le x_n \le x_N B^{n-N}$? Correct? – Q.matin Mar 05 '13 at 08:05
No, not correct. Take the $1/n$ power of both sides. – Robert Israel Mar 05 '13 at 21:36
Hmm, can you elaborate further please? – Q.matin Mar 06 '13 at 07:36
From $x_n \le x_N B^{n-N}$ you get $x_n^{1/n} \le (x_N B^{-N})^{1/n} B$. The right side has limit $B$ as $n \to \infty$. – Robert Israel Mar 06 '13 at 08:30

Proof of limit inequality

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