Prove that $\lim_{n \to\infty} \frac{x_{n+1}}{x_n} = a \implies \lim_{n \to\infty} \sqrt[n]{x_n} = a$.

Question

Given a sequence $(x_n)_{n=1}^\infty$ such that $\forall n: x_n > 0$, prove that

$$\lim_{n \to\infty} \frac{x_{n+1}}{x_n} = a \implies \lim_{n \to\infty} \sqrt[n]{x_n} = a.$$

Answer 1

The idea behind this proof is, that for $n$ sufficiently large, given that the ratio of consequent terms of the sequence is very close to $a$, we get $x_{n+1} = ax_n$, that is $x_n = a^n$. Then $\sqrt[n]{x_n} = \sqrt[n]{a^n} = a$.

Proof. Let $\epsilon > 0$. There is $n_0$ such that for all $n \ge n_0$ we have

$$a - \epsilon < \frac{a_{n+1}}{a_n} < a + \epsilon.$$

We can find bounds for $a_n$:

$$a_{n_0}(a - \epsilon)^{n - n_0} \le a_n \le a_{n_0}(a + \epsilon)^{n-n_0}.$$

Taking roots, we obtain

$$ \sqrt[n]{a_{n_0}(a - \epsilon)^{n - n_0}} \le \sqrt[n]{a_n \vphantom{()^n} } \le \sqrt[n]{a_{n_0}(a + \epsilon)^{n-n_0}}, $$ that is $$ \sqrt[n]{a_{n_0} \vphantom{()^n} }{\sqrt[n]{(a - \epsilon)^{n}}}\sqrt[n]{(a - \epsilon)^{-n_0}} \le \sqrt[n]{a_n \vphantom{()^n} } \le \sqrt[n]{a_{n_0} \vphantom{()^n} }{\sqrt[n]{(a + \epsilon)^{n}}}\sqrt[n]{(a + \epsilon)^{-n_0}}. $$

Now we have that

$$ \lim_{n \to\infty} {\sqrt[n]{a_{n_0} \vphantom{()^n} }{\sqrt[n]{(a - \epsilon)^{n}}}\sqrt[n]{(a - \epsilon)^{-n_0}}} = a - \epsilon $$

and

$$ \lim_{n \to\infty} {\sqrt[n]{a_{n_0} \vphantom{()^n} }{\sqrt[n]{(a + \epsilon)^{n}}}\sqrt[n]{(a + \epsilon)^{-n_0}}} = a + \epsilon. $$

As $\epsilon$ is arbitrary, we get that the desired limit exists and $\lim_{n \to\infty} \sqrt[n]{x_n} = a$.

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Is my proof correct? I would like to get any feedback.