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

MATH101 April 2016

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Question 02 (c)
For which values of the constant $q$ does the sum $\displaystyle \sum _{n=1}^{\infty }{\frac {(\log n)^{q}}{n}}$ converge? L: It converges only when $q<-1$. M: It converges only when $q\leq -1$. N: It converges only when $q>-1$. P: It converges only when $q\geq -1$. Q: It converges for all $q$.

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
Consider the integral test, and the conditions of using integral test

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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.