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Does the series, $\sum_{n=2}^\infty\frac{1}{(\log n)^p}$ where $p>0$ converge of diverge? Which test is suitable? Can we use comparison test?

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  • $\begingroup$ For $p=1$ You can compare the terms to the harmonic series. Since $log(n)<n$, $\frac{1}{log(n)}>\frac{1}{n}$ and the series is divergent. $\endgroup$ – aleden Nov 7 '17 at 0:30
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It is well known that for all $p>0$, $\;(\log n)^p =o(n)$, hence $$\frac1n=o\biggl(\frac1{(\log n)^p}\biggr),$$ so sinec the former diverges, the latter diverges too.

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  • $\begingroup$ What does it mean $(\log n)^p =o(n)$? $\endgroup$ – matthew Nov 7 '17 at 1:00
  • $\begingroup$ That' Landau's notation to say that $\;\dfrac{(\log n)^p}n\to 0$ when $n\to\infty$. $\endgroup$ – Bernard Nov 7 '17 at 1:08
  • $\begingroup$ But, how do we prove this limit? $\endgroup$ – matthew Nov 7 '17 at 1:51
  • $\begingroup$ Just check the log of this fraction: $\;p\log(\log n)-\log n=-\log n\biggl(1-p\dfrac{\log(\log n)}{\log n}\biggr)\to -\infty$. $\endgroup$ – Bernard Nov 7 '17 at 10:12
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We shall try the $\sum 2^ka_{2^k}$ test.

We have $$ \sum_{k=1}^\infty 2^ka_{2^k}=\sum_{k=1}^\infty \frac{2^k}{\big(\log(2^k))\big)^p} =\frac{1}{(\log 2)^p}\sum_{k=1}^\infty \frac{2^k}{k^p}. $$ The last sum is clearly divergent. (Use for example the ratio test.) Hence the original sum is also divergent.

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