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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.

babylon
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    Your reasoning doesn't actually work. You haven't justified why the limits should be equal. What do you mean by the tests being equivalent? – E.Lim Dec 02 '15 at 13:19
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    It is quite possible that one test is inconclusive and the other is not. So $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=1$ and $\lim \sqrt[n]{a_n}=x\neq 1$. Or it is possible that if $\lim a_{n+1}/a_n$ exists and $\lim\sqrt[n]{a_n}$ does not. – Thomas Andrews Dec 02 '15 at 13:21
  • $a_n$ should be strictly positive then only the assertion is true. – Kushal Bhuyan Dec 02 '15 at 13:23
  • Ok I was not confident that I was right but now I have no idea of the possible solution :/ – babylon Dec 02 '15 at 13:33
  • I do not understand how the two terms are related – babylon Dec 02 '15 at 13:34
  • Is there a way you can force out $a_n$ from the $\frac{a_{n+1}}{a_n}$ terms? – E.Lim Dec 02 '15 at 13:41
  • @ThomasAndrews $\limsup \frac{a_{n+1}}{a_n} \ge \limsup \sqrt[n]{a_n} \ge \liminf \sqrt[n]{a_n} \ge \liminf \frac{a_{n+1}}{a_n}$, so if the first exists then so does the second. – r9m Dec 02 '15 at 13:49
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    Sure, but the above argument doesn't state that. I was pointing out why the argument made by the OP is wrong, not that the theorem is wrong. @r9m – Thomas Andrews Dec 02 '15 at 13:52
  • I just realised this is actually a duplicate (oops) but thanks for your answers. http://math.stackexchange.com/questions/561076/is-lim-n-rightarrow-infty-a-n1-a-n-l-implies-lim-n-rightarrow-inft?rq=1 – babylon Dec 02 '15 at 13:56
  • @babylon) Hello , my dear, Do you know how to accept the answer ? If an answer is helpfull to you then accept the answer first and then ask the next question..No answer is accepted by you of your question.. – Empty Dec 04 '15 at 08:50

5 Answers5

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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 $$ \lim_{n\to\infty}\sqrt[\large n]{a_n}=x $$

robjohn
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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 $${\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{a_n}}} = x}.$$

Baily
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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}

Community
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Kay K.
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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

stity
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Let $b_n=\log a_n$. Then you know $\Delta b_n\to\log x$, hence $\frac 1 n\log a_n=\frac 1 n\sum_{i=0}^{n-1}\Delta b_i\to \log x$.

A.S.
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