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

First consider when $q\ge 0$.

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

By the comparison test, we have

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

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

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

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

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

$\begin{array}{rl}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{(\log x{)}^q}{x}dx& =\underset{\log 2}{\overset{\mathrm{\infty }}{\int }}{y}^{q}dy=\underset{R\to \mathrm{\infty }}{\lim }\underset{\log 2}{\overset{R}{\int }}{y}^{q}dy\\ & =\underset{R\to \mathrm{\infty }}{\lim }\frac{1}{q+1}{y}^{q+1}{|}_{\log 2}^R=\underset{R\to \mathrm{\infty }}{\lim }\frac{1}{q+1}(R^{q+1}-(\log 2{)}^{q+1})<+\mathrm{\infty },\end{array}$

provided that $q<-1$,

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

Using $\log 1=0$,

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

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

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

$\begin{array}{rl}\underset{2}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x\log x}dx& =\underset{R\to \mathrm{\infty }}{\lim }\underset{\log 2}{\overset{R}{\int }}\frac{1}{y}dy\\ & =\underset{R\to \mathrm{\infty }}{\lim }\ln y{|}_{\log 2}^R=\underset{R\to \mathrm{\infty }}{\lim }(\ln R-\log 2)=+\mathrm{\infty },\end{array}$.

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

To sum, the answer is L.