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prove or disprove this series $$\sum_{n=2}^{\infty}\dfrac{n}{\log^{10}{n}}$$ is divergent?

My idea: since $$\lim_{n\to\infty}\dfrac{\dfrac{n}{\log^{10}{n}}}{\dfrac{1}{n}}=\lim_{n\to\infty}\dfrac{n^2}{\log^{10}{n}}=\lim_{n\to\infty}\dfrac{2n^2}{10\log^9{n}}=\cdots=\infty$$ so $$\sum_{n=2}^{\infty}\dfrac{n}{\log^{10}{n}}$$ is disconverger?

if my idea is wrong.How solve this problem? if my idea true,and this problem have other methods? Thank you

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    $\begingroup$ Even simpler, the terms don't converge to $0$. $\endgroup$ – Daniel Fischer Apr 2 '14 at 11:08
  • $\begingroup$ $n=\Omega(\log^{10}(n))\implies \exists n :\frac{n}{\log^{10}(n)}\gt c\forall c\in \mathbb R$ $\endgroup$ – Guy Apr 2 '14 at 11:09
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Your idea works, since you know that $\sum \frac{1}{n}$ diverges. But it's even simpler, since $\frac{n}{\log_{10}(n)} \to +\infty$, so the series can't possibly converge.

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    $\begingroup$ I'm summing up the comments so that the question doesn't remain unanswered. $\endgroup$ – Najib Idrissi Apr 2 '14 at 11:13
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$\lim_{n \to +\infty}\frac{n}{\log_{10}(n)}=+\infty(\not =0) $ so the series diverge

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    $\begingroup$ Although it doesn't change the result, the OP's logarithm is raised to power of 10, not base 10. $\endgroup$ – Ruslan Apr 2 '14 at 11:28

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