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This question already has an answer here:

Let $\{a_n\}$ be a sequence of positive numbers. Prove that if $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} = L$$ then $$\lim_{n\rightarrow\infty}a_n^{1/n}=L.$$

The first condition means that for any $\epsilon$, then exists $N$ such that for all $n\geq N$, we have $$L-\epsilon < \dfrac{a_{n+1}}{a_n} < L+\epsilon.$$ This means $$(L-\epsilon)^na_N<a_{N+n}<(L+\epsilon)^na_N$$ for all $n\geq 0$. Then $$(L-\epsilon)^{\frac{n}{N+n}}a_N<a_{N+n}^{\frac{1}{N+n}}<(L+\epsilon)^{\frac{n}{N+n}}a_N$$

How can I finish from here?

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marked as duplicate by Martin Sleziak, Namaste, Claude Leibovici, user91500, user223391 Dec 10 '16 at 23:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You forgot to exponentiate the $a_N$; actually, your final expression is:

$$(L-\epsilon)^{n/(N+n)}a_N^{1/(N+n)}<a_{N+n}^{1(N+n)}<(L+\epsilon)^{n/(N+n)}a_N^{1/(N+n)}$$

Now by e.g. using the product rule for limits and some standard limits involving powers (note that $N$ is fixed):

$$\lim_{n\to\infty} (L-\epsilon)^{n/(N+n)}a_N^{1/(N+n)} = \lim_{n\to\infty}(L-\epsilon)^{n/(N+n)} \lim_{n\to\infty} a_N^{1/(N+n)} = (L-\epsilon)\cdot 1 = L-\epsilon$$

I trust you can conclude from here.

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In the last inequalities you have a mistake and you should write $$(L-\epsilon)^{\frac{n}{N+n}}a_N^{\frac{1}{N+n}}<a_{N+n}^{\frac{1}{N+n}}<(L+\epsilon)^{\frac{n}{N+n}}a_N^{\frac{1}{N+n}}$$ then you pass to the limit $n\to \infty$ and you find $$L-\epsilon\leq \lim_{n\to\infty}a_n^{1/n}\leq L+\epsilon$$

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  • $\begingroup$ Nice correction! $\endgroup$ – Namaste Jun 1 '14 at 12:04
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Another way to prove is to study logarithms:

$$\lim (\ln a_{n+1}-\ln a_n)=\ln L$$ Given that, our $\frac 1n \ln a_n$ is up to some tinkering an arithmetic mean of first $n$ elements of a convergent sequence $b_n=\ln a_{n+1}-\ln a_n$; it's quite easy to prove that if $x_n\to y$, then $\sum_{k=1}^n\frac{x_k}{n}\to y$, too.

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