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Let $(a_n)$ be a real sequence.

Show that $a_n>0$ and $\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=L$ implies that $\lim_{n\to\infty} a_{n}^{1/n}=L $.

Is it legitimate to use $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}=\frac{\lim_{n\to\infty} a_{n+1}}{\lim_{n\to\infty} {a_n}}=L $$

which would imply that $$\lim_{n\to\infty} a_{n+1}=L \lim_{n\to\infty} {a_n}$$

which, used recursively would lead to

$$\lim_{n\to\infty} a_{n}= \lim_{n\to\infty} L^n a_0$$

and thus using $\lim_{n\to\infty} a^{1/n}=1$ for all $a \in \mathbb{R}$ with $a>0$

$$\lim_{n\to\infty} a_{n}^{1/n}= \lim_{n\to\infty} L a_0^{1/n}=L$$

The steps I am most unsure about are the first and the last step.

B.Swan
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  • I am aware that the task is the same, however the way of solving it is my question. – B.Swan Mar 12 '17 at 18:31
  • If it is so than you may add an 'Edit" so that it is clear what you are asking for. – Jaideep Khare Mar 12 '17 at 18:33
  • I have already stated the question as it is to be asked. I have offered a solution and am asking if it is mathematically correct. – B.Swan Mar 12 '17 at 18:34
  • In my opinion, you should add an edit to inform that you know it has been asked, and what you need isn't explained in the answers of that question. – Jaideep Khare Mar 12 '17 at 18:37
  • It is not. The limit $\lim_{n\rightarrow\infty}a_n$ is (usually) either $0$ or $\infty$, so saying $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}L^na_0$ is not saying anything. – Bob Jones Mar 12 '17 at 18:37

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