# Convergence of $\sum_{n=2}^\infty \frac {1}{(\log n )^3}$

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.

• For some sufficiently large $n$ one has : $$n\gt(\log n)^3,$$ hence $$\dfrac1n\lt\dfrac1{(\log n)^3}.$$ – Workaholic Jan 22 '15 at 20:48
• More general case: math.stackexchange.com/q/649933 – Jonas Meyer Jan 22 '15 at 20:52
• @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. – Workaholic Jan 22 '15 at 20:57
• @Wanderer You're welcome. – Workaholic Jan 22 '15 at 20:59
• Typo: It should be $n^2\gt(\log n)^3$ in my second comment. – Workaholic Jan 22 '15 at 21:09

HINT: $\log x<x^{1/4}$ for large $x$.