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

MATH101 April 2018

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[hide] 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. ${\textstyle \displaystyle \sum _{n=2}^{\infty }{\frac {1}{n(\log n)^{3}}}}$ ${\textstyle \displaystyle \sum _{n=0}^{\infty }{\frac {(-1)^{n}4^{n+1}}{(2^{2n}+1)}}}$ ${\textstyle \displaystyle \sum _{n=0}^{\infty }(-1)^{n}{\frac {\cos(n)}{n^{2}}}}$ ${\textstyle \displaystyle \sum _{n=1}^{\infty }{\frac {4e^{n}}{n!}}}$

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

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[show] Solution
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 ${\textstyle {\color {blue}T,S,O,N}}$ .