# Science:Math Exam Resources/Courses/MATH101/April 2016/Question 02 (c)/Solution 1

First consider when ${\displaystyle q\geq 0}$.

For ${\displaystyle n\geq 3}$, we have ${\displaystyle \log n\geq 1}$, which implies ${\displaystyle (\log n)^{q}\geq 1}$ and ${\displaystyle {\frac {(\log n)^{q}}{n}}>{\frac {1}{n}}}$.

By the comparison test, we have

${\displaystyle \sum _{n=1}^{\infty }{\frac {(\log n)^{q}}{n}}\geq \sum _{n=3}^{\infty }{\frac {(\log n)^{q}}{n}}\geq \sum _{n=3}^{\infty }{\frac {1}{n}}=+\infty .}$

Therefore, for ${\displaystyle q\geq 0}$, the given series diverges.

On the other hand, for ${\displaystyle q<0}$, note that ${\displaystyle {\frac {(\log x)^{q}}{x}}}$ is continuous, positive, and decreasing on ${\displaystyle [2,\infty )}$.

In this case, we'll use the integral test.

For ${\displaystyle q<0}$ but ${\displaystyle q\neq -1}$, we have

{\displaystyle {\begin{aligned}\int _{2}^{\infty }{\frac {(\log x)^{q}}{x}}dx&=\int _{\log 2}^{\infty }y^{q}dy=\lim _{R\to \infty }\int _{\log 2}^{R}y^{q}dy\\&=\lim _{R\to \infty }{\frac {1}{q+1}}y^{q+1}{\bigg |}_{\log 2}^{R}=\lim _{R\to \infty }{\frac {1}{q+1}}\left(R^{q+1}-(\log 2)^{q+1}\right)<+\infty ,\end{aligned}}}

provided that ${\displaystyle q<-1}$,

Here, the second equality is obtained from the substitution ${\displaystyle y=\log x}$.

Using ${\displaystyle \log 1=0}$,

${\displaystyle \sum _{n=1}^{\infty }{\frac {(\log n)^{q}}{n}}=\sum _{n=2}^{\infty }{\frac {(\log n)^{q}}{n}}.}$

Then, by the integral test, the given series is convergent for ${\displaystyle q<-1}$; ,

For ${\displaystyle q=-1}$, the integral can be computed by

{\displaystyle {\begin{aligned}\int _{2}^{\infty }{\frac {1}{x\log x}}dx&=\lim _{R\to \infty }\int _{\log 2}^{R}{\frac {1}{y}}dy\\&=\lim _{R\to \infty }\ln y{\bigg |}_{\log 2}^{R}=\lim _{R\to \infty }\left(\ln R-\log 2\right)=+\infty ,\end{aligned}}}.

Therefore, again by the integral test, the given series diverges when ${\displaystyle q=-1}$.

To sum, the answer is L.