Science:Math Exam Resources/Courses/MATH103/April 2014/Question 06 (c)

MATH103 April 2014

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[hide] Question 06 (c)
Determine with full justification whether the following series converge. You do not have to evaluate the sums. $\displaystyle \sum _{k=2}^{\infty }{\frac {1}{k(\ln k)^{103}}}$ Work must be shown for full marks. Simplify fully.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Use the integral test.

[show] Hint 2
If your first attempt at using the integral test yielded divergence, it is probably because the interval on which you are integrating is not well-chosen. Precisely, it is likely that the integrand is not well-defined and/or monotone increasing on this interval (and this is required in order for the integral test to give the correct result). Change your bounds of integration and try again.

Checking a solution serves two purposes: helping you if, after having used all the hints, 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.
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[show] 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. Use the integral test. That is, we must integrate the function $f(x)={\frac {1}{x(\ln x)^{103}}}$ over an interval $[a,\infty )$ where it is well-defined and monotone decreasing. Since $f(x)$ is undefined at $x=1$ , we must choose $a>1$ . For any such choice of $a$ , $f(x)$ will be well-defined and decreasing on the resulting interval $[a,\infty )$ . For instance, choose $a=2$ and consider the integral $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{103}}}\mathrm {d} x.$ To evaluate this integral, use substitution. Let $u=\ln x$ , so $\mathrm {d} u={\frac {1}{x}}\mathrm {d} x$ . Note that since $x$ ranges from $2$ to $\infty$ , the variable $u=\ln x$ ranges from $\ln 2$ to $\infty$ . Thus, $\int _{2}^{\infty }{\frac {1}{x(\ln x)^{103}}}\mathrm {d} x=\int _{\ln 2}^{\infty }{\frac {1}{u^{-103}}}\mathrm {d} u=\left.{\frac {1}{-102}}u^{-102}\right\|_{\ln 2}^{\infty }={\frac {1}{102}}(\ln 2)^{-102}<\infty$ and it follows that the corresponding sum $\sum _{k=1}^{\infty }{\frac {1}{k(\ln k)^{103}}}$ converges.