Consider a series $\sum_{n \geq 0} a_n$. How can I show the following?
$$\lim_{n \to \infty} |a_n|^{\frac{1}{n}}=l \implies \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}=l$$
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Martin Sleziak
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Gianolepo
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It works in the opposite direction. See, for example Convergence of Ratio Test implies Convergence of the Root Test and the posts linked there. – Martin Sleziak Dec 05 '16 at 23:44
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You can't, because it's not true. For example, take $$a_n=\cases{4^n&if $n$ is even\cr2\times4^n\quad&if $n$ is odd.\cr}$$ Then $$a_n^{1/n}=\cases{4&if $n$ is even\cr4\times2^{1/n}\quad&if $n$ is odd\cr}$$ which has limit $l=4$. But $$\frac{a_{n+1}}{a_n}=\cases{8&if $n$ is even\cr2\quad&if $n$ is odd\cr}$$ which has no limit.
David
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Thanks for the counter example. Could I prove the statement if I suppose in advance that both $\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}$ and $\lim_{n \to \infty} |a_n|^{1/n}$ do exist? – Gianolepo Dec 06 '16 at 16:06
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Take $a_n=0$
$$\lim_{n\to +\infty} |a_n|^{\frac{1}{n}}=0$$
but
$\lim_{n\to +\infty}\frac{|a_{n+1}|}{|a_n|}$ is indeterminate.
hamam_Abdallah
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