Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 13/Solution 1

Since $q>0$, the dominant term in $(3n)^{q}-1$ is $(3n)^{q}$. Therefore, we would like to compare the series to $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {(3n)^{q}}}}=\sum _{n=1}^{\infty }(3n)^{-q/2}$. In order to justify this, we must use the Limit Comparison Test. If $a_{n}={\frac {1}{\sqrt {(3n)^{q}-1}}}$ and $b_{n}={\frac {1}{\sqrt {(3n)^{q}}}}$, then

$${\begin{aligned}\lim _{n\to \infty }{\frac {b_{n}}{a_{n}}}=\lim _{n\to \infty }{\frac {1/{\sqrt {(3n)^{q}}}}{1/{\sqrt {(3n)^{q}-1}}}}=\lim _{n\to \infty }{\frac {\sqrt {(3n)^{q}-1}}{\sqrt {(3n)^{q}}}}=\lim _{n\to \infty }{\sqrt {1-{\frac {1}{(3n)^{q}}}}}=1.\end{aligned}}$$

Therefore, $\sum _{n=1}^{\infty }a_{n}$ converges if and only if $\sum _{n=1}^{\infty }b_{n}$ converges. So let us figure out which values of $q$ make $\sum _{n=1}^{\infty }b_{n}$ converge.

Simplifying, $\sum _{n=1}^{\infty }b_{n}=\sum _{n=1}^{\infty }(3n)^{-q/2}=3^{-q/2}\sum _{n=1}^{\infty }n^{-q/2}$. By the p-test, this converges for $q>2$, and diverges for $q\leq 2$.

In conclusion, the original series $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {(3n)^{q}-1}}}$ converges if and only if $q>2$.