Science:Math Exam Resources/Courses/MATH105/April 2015/Question 05 (d)

MATH105 April 2015

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Question 05 (d)
Determine whether the series $\sum \limits _{k=2}^{\infty }{\frac {1}{k(\ln k)^{3}}}$ converges or diverges.

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
The function ${\frac {1}{x(\ln x)^{3}}}$ appears in the series and has a closed-form antiderivative. This means that applying the integral test may be helpful here.

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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. We will apply the integral test. The test states that if the integral $\int _{n}^{\infty }f(x)\,dx$ converges, the series $\sum _{k=n}^{\infty }f(k)$ converges as well. We may evaluate the integral $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{3}}}\,dx$ by a substitution. Let $u=\ln x$. Then $du={\frac {dx}{x}}$, which implies that $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{3}}}\,dx=\int _{\ln 2}^{\infty }{\frac {du}{u^{3}}}=\lim _{b\to \infty }\left.-{\frac {1}{2u^{2}}}\right|_{\ln 2}^{b}=\lim _{b\rightarrow \infty }\left(-{\frac {1}{2b^{2}}}+{\frac {1}{2(\ln 2)^{2}}}\right)={\frac {1}{2(\ln 2)^{2}}}.$ Therefore, the integral $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{3}}}\,dx$ converges, and therefore by the integral test, the series $\sum _{k=2}^{\infty }{\frac {1}{k(\ln k)^{3}}}$ converges also.