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Test Convergence of $\displaystyle\sum_{n=2}^\infty \dfrac {1}{(\log n )^3}$

Attempt: I haven't been able to find a suitable comparator for $\dfrac {1}{(\log n )^3}$ . The integration test also seems tedious.

Please guide me on how to move forward.

Thank you very much for your help.

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    $\begingroup$ For some sufficiently large $n$ one has : $$ n\gt(\log n)^3, $$ hence $$ \dfrac1n\lt\dfrac1{(\log n)^3}. $$ $\endgroup$ – Workaholic Jan 22 '15 at 20:48
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    $\begingroup$ More general case: math.stackexchange.com/q/649933 $\endgroup$ – Jonas Meyer Jan 22 '15 at 20:52
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    $\begingroup$ @Wanderer When I first saw the series, I immediately thought of $\sum\limits_{n=1}^\infty\frac1n$, so I tried to find whether $\frac1n\lt\frac1{(\log n)^3}$ holds. But with $n^2\lt(\log n)^3$ we will most probably get nothing since $\sum\limits_{n=1}^\infty\frac1{n^2}$ converges. $\endgroup$ – Workaholic Jan 22 '15 at 20:57
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    $\begingroup$ @Wanderer You're welcome. $\endgroup$ – Workaholic Jan 22 '15 at 20:59
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    $\begingroup$ Typo: It should be $n^2\gt(\log n)^3$ in my second comment. $\endgroup$ – Workaholic Jan 22 '15 at 21:09
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HINT: $\log x<x^{1/4}$ for large $x$.

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