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I am having trouble determining convergence/ divergence of the following series:

$\sum\limits_{n=2}^{\infty} \frac{1}{(\log(n))^{1/n}}$

When I apply the root and ratio tests, I find in both cases the limit 1 (which means that the test is inconclusive).

Please help

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Hint: What is $\lim\limits_{n\rightarrow\infty} \bigl(\log (n)\bigr)^{1/n}$? – David Mitra Dec 12 '12 at 13:18

Just use the limit test. It is $\log(n)^{1/n}\longrightarrow 1\neq 0$, so the series cannot converge.

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The limit test states that if $\lim_{n \to \infty}\frac{a_n}{b_n}=1$ and $\sum b_n$ converges then $\sum a_n$ converges. How did you use it here? – Carpediem Dec 12 '12 at 13:41
@user43758 He is not referring to that test, but rather the following: If $\lim_{n\rightarrow\infty} a_n\ne 0$, then the series $\sum_{n=1}^\infty a_n$ diverges. – David Mitra Dec 12 '12 at 13:43
@DavidMitra note that's actually the contrapositive of the usual theorem that's taught ($\sum a_n$ converges $\implies a_n \to 0$). – DanZimm Jun 2 '14 at 9:44

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