I know that, if $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = x$, then $\lim_{n\to\infty} |a_n|^\frac{1}{n} = x$. However, does $\lim_{n\to\infty} |a_n|^\frac{1}{n} = x$ and $a_n \neq 0$ imply $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = x$? I can't seem to find a counterexample nor do I know how to prove that it is true. Can someone help me please? Thanks.
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$\begingroup$ Wikipedia mentions an example where ratio test is inconclusive but root test works. It is obtained by "doubling" a convergent series, i.e., every term is taken twice. (But I am sure that other interesting examples are also possible. Here is a bunch of results from Google Books where some other examples could be found.) $\endgroup$– Martin SleziakMay 2, 2014 at 15:07
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Suppose $a_n = (-1)^n$. then $|a_n|^{1/n} = 1 \to 1 = x$. However
$$ \frac{ a_{n+1}}{a_n} = \frac{ (-1)^{n+1}}{(-1)^n} = -1 \to -1 \neq x$$
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$\begingroup$ Hi just wondering, what if a_n are all positive, can counterexample still be found? $\endgroup$– 123May 2, 2014 at 14:38