Prove that if $\lim_{n\to \infty}{\frac{a_{n+1}}{a_n}}=x$ then $\lim_{n\to \infty}{\sqrt[n]{a_n}}=x$ Prove that if $\lim_{n\to \infty}{\frac{a_{n+1}}{a_n}}=x$ then $\lim_{n\to \infty}{\sqrt[n]{a_n}}=x$
My proposed solution uses the following prepositions:
Proposition 4.7. Let $a_n$ be a sequence of real numbers such that ${\sqrt[n]{a_n}}$ converges to L. If L < 1
the sequence converges to zero, if L > 1 the sequence is divergent, if L = 1 the test is inconclusive.
Proposition 4.8. Let $a_n$ be a sequence of real numbers such that $\frac{a_{n+1}}{a_n}$ converges to L.
If L < 1 the sequence converges to zero, if L > 1 the sequence is divergent, if L = 1 the test is inconclusive.
These tests are perfectly equivalent and so their limits must be the same.
That is my solution but we were given the hint that we could use the result $\lim_{n\to \infty}{a_n^s}=x^s$ where s is rational and I have not used this hint which makes me think my solution is wrong. Also my solution seems too simple.
 A: Take the log of
$$
\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=x
$$
and rewrite it as
$$
\lim_{n\to\infty}\frac{\log(a_{n+1})-\log(a_n)}{(n+1)-n}=\log(x)
$$
Then apply the Stolz–Cesàro theorem to get
$$
\lim_{n\to\infty}\frac{\log(a_n)}{n}=\log(x)
$$
Apply $e^x$ to get
$$
\lim_{n\to\infty}\sqrt[\large n]{a_n}=x
$$
A: Let $\varepsilon>0$. Since $\frac{a_{n+1}}{a_n}\to x$ as $n\to\infty$ there exists a $n_0\in\mathbb{N}$ such that $\left| {\frac{{{a_{n + 1}}}}{{{a_n}}} - x} \right| \le \varepsilon $ for all $n\ge n_0$. This leads to
$$\left( {x - \varepsilon } \right){a_n} \le {a_{n + 1}} \le \left( {x + \varepsilon } \right){a_n},\,\,\,\forall n \geqslant {n_0}.$$
By induction, we get
$${\left( {x - \varepsilon } \right)^{n - {n_0}}}{a_{{n_0}}} \leqslant {a_n} \leqslant {\left( {x + \varepsilon } \right)^{n - {n_0}}}{a_{{n_0}}},\,\,\,\,\forall n \geqslant {n_0}$$
$$\Rightarrow {\left( {x - \varepsilon } \right)^{\frac{{n - {n_0}}}
{n}}}{\left( {{a_{{n_0}}}} \right)^{\frac{1}{n}}} \leqslant \sqrt[n]{{{a_n}}} \leqslant {\left( {x + \varepsilon } \right)^{\frac{{n - {n_0}}}{n}}}{\left( {{a_{{n_0}}}} \right)^{\frac{1}{n}}},\,\,\,\,\forall n \geqslant {n_0}.$$
Let $n\to\infty$, we get
$$x - \varepsilon  \leqslant \mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} \leqslant x + \varepsilon  \Rightarrow \left| {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} - x} \right| \leqslant \varepsilon .$$
So,
$$\left| {\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} - x} \right| = 0$$
or
$${\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = x}.$$
A: We need another theorem like the Stolz–Cesàro theorem in the other answer or the one here If $\lim a_n = L$, then $\lim s_n = L$ that I will use here.
\begin{align}
&Let\space b_n = \ln{\frac{a_{n+1}}{a_n}}\\
&\lim_{n\to\infty} b_n = \ln x\\
&\lim_{n\to\infty} \frac{\sum_{k=1}^n b_k}{n} =\lim_{n\to\infty} \frac{\ln a_{n+1}-\ln a_1}{n}=\ln x\\
&\lim_{n\to\infty} \frac{\ln a_{n}}{n}=\ln x\\
&\lim_{n\to\infty} e^{\frac{\ln a_{n}}{n}}=\lim_{n\to\infty} \sqrt[n]{a_n}= x
\end{align}
A: Hint : try to prove that $$\forall k , \lim_{n \to \infty} \frac {a_{n+1}}{x^{n+1-k}a_k}=1$$ then use $k=0$ to conclude
A: Let $b_n=\log a_n$. Then you know $\Delta b_n\to\log x$, hence $\frac 1 n\log a_n=\frac 1 n\sum_{i=0}^{n-1}\Delta b_i\to \log x$.
