Science:Math Exam Resources/Courses/MATH101 A/April 2024/Question 13

MATH101 A April 2024

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Question 13
Find the set of all $q>0$ for which the following series converges: $${\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{\sqrt {(3n)^{q}-1}}}.\end{aligned}}$$ As in all questions, remember to fully justify your answer.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
In the expression $(3n)^{q}-1$, what is the dominant term? If we ignore everything but the dominant term, we should get a series whose convergence or divergence is easily verified by the p-Test (CLP Example 3.3.6). We can use a comparison test to show that the original series behaves the same way.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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 $q>2$.