Science:Math Exam Resources/Courses/MATH101/April 2018/Question 02 (iii)

MATH101 April 2018

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Question 02 (iii)
For each of the following series, choose the appropriate statement. Write N, O, P, S, or T in each box. Each answer will be used at most once, and each series matches a single answer only. N: The series converges by the Ratio Test. O: The series converges absolutely by the Comparison Test with a ${\textstyle p}$ -series. P: The series converges by the Alternating Series Test. S: The series diverges. T: The series converges by the Integral Test. ${\textstyle \displaystyle \sum _{n=2}^{\infty }{\frac {1}{n(\log n)^{3}}}}$ ${\textstyle \displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}4^{n+1}}{(2^{2n}+1)}}}$ ${\textstyle \displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {\cos(n)}{n^{2}}}}$ ${\textstyle \displaystyle \sum _{n=1}^{\infty }{\frac {4e^{n}}{n!}}}$

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
The easiest check for divergence of a series is if the summand does not tend to zero.

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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 relevant integral for the first series is ${\textstyle \int _{2}^{\infty }{\frac {1}{x(\log x)^{3}}}dx}$ which can be computed using the substitution ${\textstyle u=\log x}$ and shows that the series converges by the Integral Test. For the second series, observe that we have $\lim _{n\to \infty }{\frac {(-1)^{n}4^{n+1}}{2^{2n}+1}}=\lim _{n\to \infty }{\frac {4\cdot (-1)^{n}\cdot 4^{n}}{4^{n}+1}}=\lim _{n\to \infty }{\frac {4(-1)^{n}}{1+{\frac {1}{4^{n}}}}}=\lim _{n\to \infty }4(-1)^{n},$ so the even terms of the limit tend to 4 and the odd terms to -4, which shows that the summand does not tend to zero (the limit does not even exist). This means that the series $\sum _{n=0}^{\infty }{\frac {(-1)^{n}4^{n+1}}{(2^{2n}+1)}}$ diverges. For the third series, for any natural number ${\textstyle n}$ , we have $\left\|(-1)^{n}{\frac {\cos(n)}{n^{2}}}\right\|=\left\|{\frac {\cos(n)}{n^{2}}}\right\|\leq {\frac {1}{n^{2}}},$ so the series converges absolutely by the Comparison Test with a ${\textstyle p}$ -series. For the fourth series, the Ratio Test gives $\lim _{n\to \infty }\left\|{\frac {\frac {4e^{n+1}}{(n+1)!}}{\frac {4e^{n}}{n!}}}\right\|=\lim _{n\to \infty }\left\|{\frac {4e^{n+1}}{(n+1)!}}{\frac {n!}{4e^{n}}}\right\|=\lim _{n\to \infty }{\frac {4e}{n+1}}=0,$ which shows that the series converges. Answer: The respective answers are ${\textstyle {\color {blue}T,S,O,N}}$ .