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

First consider when $q\geq 0$.

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

By the comparison test, we have

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

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

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

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

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

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

provided that $q<-1$,

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

Using $\log 1=0$,

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

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

For $q=-1$, the integral can be computed by

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

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

To sum, the answer is L.