Science:Math Exam Resources/Courses/MATH101/April 2015/Question 02 (c) (i)

MATH101 April 2015

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Question 02 (c) (i)

For each of the following series, choose the appropriate statement. (Write the appropriate two-letter answer in each box; each answer will be used at most once.) CI: Converges by Integral Test with $\int _{1}^{\infty }{\frac {\ln x}{x}}dx$ CT: Converges by Test for Divergence CC: Converges by Comparison Test with $\sum _{n=1}^{\infty }{\frac {1}{n^{3}}}$ CA: Converges by Alternating Series Test CL: Converges by Limit Comparison Test with $\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$ DI: Diverges by Integral Test with $\int _{1}^{\infty }{\frac {\ln x}{x}}dx$ DT: Diverges by Test for Divergence DC: Diverges by Comparison Test with $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}$ DA: Diverges by Alternating Series Test DL: Diverges by Limit Comparison Test with $\sum _{n=1}^{\infty }{\frac {\pi }{n^{2}}}$ (i) $\sum _{n=1}^{\infty }{\frac {\ln n}{n}}$

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
Note that the function ${\frac {\ln x}{x}}$ is decreasing for $x>e$, and what we really care for is large n's.

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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. The related function to the summand is $f(x)={\frac {\ln x}{x}}$, which is non-negative for $x\geq 1$, so $f(x)\geq 0$ . Next, we check that $f(x)={\frac {\ln x}{x}}$ is decreasing. We show that the derivative is negative. $f'(x)={\frac {{{\frac {1}{x}}\cdot x}-\ln x}{x^{2}}}={\frac {1-\ln x}{x^{2}}}$. $\ln x>1$ when $x>e$, so as $n$ get larger we will have $\ln x>1$ and therefore, ${\frac {1-\ln x}{x^{2}}}\leq 0$ which means that the function is decreasing, so we can apply the integral test. Now we compute the integral corresponding to the series $\int {\frac {\ln x}{x}}\,dx$: We apply the following substitution: $\ln x=u,\ {\frac {dx}{x}}=du$ $\int _{1}^{\infty }{\frac {\ln x}{x}}\,dx=\int _{0}^{\infty }u\,du={\frac {1}{2}}u^{2}\vert _{0}^{\infty }=\infty$ The integral diverges so the series diverges as well. Answer: DI