# About the convergence of a real series

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?

• 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. – aleden Nov 7 '17 at 0:30

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.
• What does it mean $(\log n)^p =o(n)$? – matthew Nov 7 '17 at 1:00
• That' Landau's notation to say that $\;\dfrac{(\log n)^p}n\to 0$ when $n\to\infty$. – Bernard Nov 7 '17 at 1:08
• Just check the log of this fraction: $\;p\log(\log n)-\log n=-\log n\biggl(1-p\dfrac{\log(\log n)}{\log n}\biggr)\to -\infty$. – Bernard Nov 7 '17 at 10:12
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.