此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。
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本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution
Summary
- If $$sum_{n=1}^infty |a_n|$$ converges (i.e. absolutely convergent), then $$sum_{n=1}^infty a_n$$ converges (i.e. conditionally convergent).
- Suppose that $(a_n)$ is a decreasing sequence of positive numbers and $$lim_{n oinfty}a_n=0$$ Then the alternating series $$sum_{n=1}^infty (-1)^{n+1} a_n$$ converges.
- For an alternating series $$s_n=sum_{n=1}^{infty}(-1)^ncdot a_n$$ the test steps:
- If $$lim_{n oinfty}a_n eq0$$ then it diverges;
- If $$lim_{n oinfty}a_n=0$$ and $a_n$ converges, then it absolutely converges;
- If $$lim_{n oinfty}a_n=0$$ and $a_n$ diverges, then it conditionally converges.
Exercises 4.1
Determine whether each series converges absolutely, converges conditionally, or diverges.
1. $$sum_{n=1}^infty (-1)^{n-1}{1over 2n^2+3n+5}$$ Solution: $$sum_{n=1}^{infty}|a_n|=sum_{n=1}^{infty}{1over 2n^2+3n+5} < sum_{n=1}^{infty}{1over 2n^2} o ext{converge}$$ Thus it converges absolutely.
2. $$sum_{n=1}^infty (-1)^{n-1}{3n^2+4over 2n^2+3n+5}$$ Solution: $$lim_{n oinfty}|a_n|=lim_{n oinfty}{3n^2+4 over 2n^2+3n+5}={3over2} eq0$$ Thus it diverges.
3. $$sum_{n=1}^infty (-1)^{n-1}{ln nover n}$$ Solution: $$lim_{n oinfty}{ln nover n}=0$$ and $${ln nover n} > {1over n} o ext{diverge}$$ Thus it converges conditionally.
4. $$sum_{n=1}^infty (-1)^{n-1} {ln nover n^3}$$ Solution: $$lim_{n oinfty}{ln nover n^3}=0$$ and $${ln nover n^3} < {nover n^3}={1over n^2} o ext{converge}$$ Thus it converges absolutely.
5. $$sum_{n=2}^infty (-1)^n{1over ln n}$$ Solution: $$lim_{n oinfty}{1overln n}=0$$ and $${1overln n} > {1over n} o ext{diverge}$$ Thus it converges conditionally.
6. $$sum_{n=0}^infty (-1)^{n} {3^nover 2^n+5^n}$$ Solution: $$lim_{n oinfty}{3^nover 2^n+5^n}=0$$ and $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{3^{n+1}over 2^{n+1}+5^{n+1}}cdot{2^n+5^nover 3^n}$$ $$=lim_{n oinfty}{3cdot(2^n+5^n)over 2^{n+1}+5^{n+1}}={3over5} < 1$$ Thus it converges absolutely.
7. $$sum_{n=0}^infty (-1)^{n} {3^nover 2^n+3^n}$$ Solution: $$lim_{n oinfty}{3^nover 2^n+3^n}=1 eq0$$ Thus it diverges.
8. $$sum_{n=1}^infty (-1)^{n-1} {arctan nover n}$$ Solution: $$lim_{n oinfty}{arctan nover n}=lim_{n oinfty}{1over 1+n^2}=0$$ and $${arctan nover n} > {1over n} o ext{diverge}$$ Thus it converges conditionally.
Exercises 4.2
Determine whether the following series converge or diverge.
1. $$sum_{n=1}^infty {(-1)^{n+1}over 2n+5}$$ Solution: $$lim_{n oinfty}{1over 2n+5}=0$$ Thus it converges.
2. $$sum_{n=4}^infty {(-1)^{n+1}over sqrt{n-3}}$$ Solution: $$lim_{n oinfty}{1over sqrt{n-3}}=0$$ Thus it converges.
3. $$sum_{n=1}^infty (-1)^{n+1}{nover 3n-2}$$ Solution: $$lim_{n oinfty}{nover 3n-2}={1over3} eq0$$ Thus it diverges.
4. $$sum_{n=1}^infty (-1)^{n+1}{ln nover n}$$ Solution: $$lim_{n oinfty}{ln nover n}=0$$ Thus it converges.
5. Approximate $$sum_{n=1}^infty (-1)^{n+1}{1over n^3}$$ to two decimal places.Solution: $$int_{N}^{infty}{1over x^3}dx= -{1over2}cdot{1over x^2}Big|_{N}^{infty}= {1over2}cdot{1over N^2} < {1over100}Rightarrow N geq 8$$ Adding up the first 8 terms and the result is $0.9007447doteq0.90$.
6. Approximate $$sum_{n=1}^infty (-1)^{n+1}{1over n^4}$$ to two decimal places.Solution: $$int_{N}^{infty}{1over x ^4}dx=-{1over3}cdot{1over x^3}Big|_{N}^{infty}={1over3}cdot{1over N^3} < {1over100}Rightarrow Ngeq4$$ Adding up the first 4 term and the result is $0.9459394doteq0.95$.
Additional Exercises
1. Suppose $$sum_{n=1}^{infty}|a_n|$$ converges, what about $$sum_{n=1}^{infty}a_n$$ Solution: $$sum_{n=1}^{infty}|a_n| ext{converges}$$ $$Rightarrowsum_{n=1}^{infty}2cdot|a_n| ext{converges}$$ We have $$0leq a_n+|a_n|leq2cdot|a_n|$$ By comparison test, $$sum_{n=1}^{infty}(a_n+|a_n|)$$ converges. And $$sum_{n=1}^{infty}(a_n+|a_n|)-sum_{n=1}^{infty}|a_n|=sum_{n=1}^{infty}a_n$$ converges.
This exercise shows that "Absolutely converge implies converge".
2. $$sum_{j=5}^{infty}{2j^2+j+2 over 3j^5+j^4+5j^3+6}$$ converge or diverge?
Solution: $${2j^2+j+2 over 3j^5+j^4+5j^3+6} < {3i^2over3j^5}={1over j^3} o ext{converge}$$ By $p$-series test and comparison test, it converges.
3. $$sum_{n=2}^{infty}{6cdot(-1)^n over 7n^{0.52}}$$ converge or diverge?
Solution: $$lim_{n oinfty}{6over 7n^{0.52}}=0$$ and $${6over 7n^{0.52}} > {1over 7n^{0.52}} o ext{diverge}$$ Thus it converges conditionally.
4. $$sum_{n=7}^{infty}{4cdot(-1)^{n+1}over n^2+3n+5}$$ converge or diverge?
Solution: $$lim_{n oinfty}{4over n^2+3n+5}=0$$ and $${4over n^2+3n+5} < {4over n^2} o ext{converge}$$ Thus it converges absolutely.