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 $$\lim_{n\to\infty}a_n=0$$ Then the alternating series $$\sum_{n=1}^\infty (-1)^{n+1} a_n$$ converges.
- For an alternating series $$s_n=\sum_{n=1}^{\infty}(-1)^n\cdot a_n$$ the test steps:
- If $$\lim_{n\to\infty}a_n\neq0$$ then it diverges;
- If $$\lim_{n\to\infty}a_n=0$$ and $a_n$ converges, then it absolutely converges;
- If $$\lim_{n\to\infty}a_n=0$$ and $a_n$ diverges, then it conditionally converges.
Exercises 4.1
Determine whether each series converges absolutely, converges conditionally, or diverges.
1. $$\sum_{n=1}^\infty (-1)^{n-1}{1\over 2n^2+3n+5}$$ Solution: $$\sum_{n=1}^{\infty}|a_n|=\sum_{n=1}^{\infty}{1\over 2n^2+3n+5} < \sum_{n=1}^{\infty}{1\over 2n^2}\to\text{converge}$$ Thus it converges absolutely.
2. $$\sum_{n=1}^\infty (-1)^{n-1}{3n^2+4\over 2n^2+3n+5}$$ Solution: $$\lim_{n\to\infty}|a_n|=\lim_{n\to\infty}{3n^2+4 \over 2n^2+3n+5}={3\over2}\neq0$$ Thus it diverges.
3. $$\sum_{n=1}^\infty (-1)^{n-1}{\ln n\over n}$$ Solution: $$\lim_{n\to\infty}{\ln n\over n}=0$$ and $${\ln n\over n} > {1\over n}\to\text{diverge}$$ Thus it converges conditionally.
4. $$\sum_{n=1}^\infty (-1)^{n-1} {\ln n\over n^3}$$ Solution: $$\lim_{n\to\infty}{\ln n\over n^3}=0$$ and $${\ln n\over n^3} < {n\over n^3}={1\over n^2}\to\text{converge}$$ Thus it converges absolutely.
5. $$\sum_{n=2}^\infty (-1)^n{1\over \ln n}$$ Solution: $$\lim_{n\to\infty}{1\over\ln n}=0$$ and $${1\over\ln n} > {1\over n}\to\text{diverge}$$ Thus it converges conditionally.
6. $$\sum_{n=0}^\infty (-1)^{n} {3^n\over 2^n+5^n}$$ Solution: $$\lim_{n\to\infty}{3^n\over 2^n+5^n}=0$$ and $$\lim_{n\to\infty}a_{n+1}/a_n=\lim_{n\to\infty}{3^{n+1}\over 2^{n+1}+5^{n+1}}\cdot{2^n+5^n\over 3^n}$$ $$=\lim_{n\to\infty}{3\cdot(2^n+5^n)\over 2^{n+1}+5^{n+1}}={3\over5} < 1$$ Thus it converges absolutely.
7. $$\sum_{n=0}^\infty (-1)^{n} {3^n\over 2^n+3^n}$$ Solution: $$\lim_{n\to\infty}{3^n\over 2^n+3^n}=1\neq0$$ Thus it diverges.
8. $$\sum_{n=1}^\infty (-1)^{n-1} {\arctan n\over n}$$ Solution: $$\lim_{n\to\infty}{\arctan n\over n}=\lim_{n\to\infty}{1\over 1+n^2}=0$$ and $${\arctan n\over n} > {1\over n}\to\text{diverge}$$ 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: $$\lim_{n\to\infty}{1\over 2n+5}=0$$ Thus it converges.
2. $$\sum_{n=4}^\infty {(-1)^{n+1}\over \sqrt{n-3}}$$ Solution: $$\lim_{n\to\infty}{1\over \sqrt{n-3}}=0$$ Thus it converges.
3. $$\sum_{n=1}^\infty (-1)^{n+1}{n\over 3n-2}$$ Solution: $$\lim_{n\to\infty}{n\over 3n-2}={1\over3}\neq0$$ Thus it diverges.
4. $$\sum_{n=1}^\infty (-1)^{n+1}{\ln n\over n}$$ Solution: $$\lim_{n\to\infty}{\ln n\over n}=0$$ Thus it converges.
5. Approximate $$\sum_{n=1}^\infty (-1)^{n+1}{1\over n^3}$$ to two decimal places.Solution: $$\int_{N}^{\infty}{1\over x^3}dx= -{1\over2}\cdot{1\over x^2}\Big|_{N}^{\infty}= {1\over2}\cdot{1\over N^2} < {1\over100}\Rightarrow N \geq 8$$ Adding up the first 8 terms and the result is $0.9007447\doteq0.90$.
6. Approximate $$\sum_{n=1}^\infty (-1)^{n+1}{1\over n^4}$$ to two decimal places.Solution: $$\int_{N}^{\infty}{1\over x ^4}dx=-{1\over3}\cdot{1\over x^3}\Big|_{N}^{\infty}={1\over3}\cdot{1\over N^3} < {1\over100}\Rightarrow N\geq4$$ Adding up the first 4 term and the result is $0.9459394\doteq0.95$.
Additional Exercises
1. Suppose $$\sum_{n=1}^{\infty}|a_n|$$ converges, what about $$\sum_{n=1}^{\infty}a_n$$ Solution: $$\sum_{n=1}^{\infty}|a_n|\ \text{converges}$$ $$\Rightarrow\sum_{n=1}^{\infty}2\cdot|a_n|\ \text{converges}$$ We have $$0\leq a_n+|a_n|\leq2\cdot|a_n|$$ 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: $${2j^2+j+2 \over 3j^5+j^4+5j^3+6} < {3i^2\over3j^5}={1\over j^3}\to\text{converge}$$ By $p$-series test and comparison test, it converges.
3. $$\sum_{n=2}^{\infty}{6\cdot(-1)^n \over 7n^{0.52}}$$ converge or diverge?
Solution: $$\lim_{n\to\infty}{6\over 7n^{0.52}}=0$$ and $${6\over 7n^{0.52}} > {1\over 7n^{0.52}}\to\text{diverge}$$ Thus it converges conditionally.
4. $$\sum_{n=7}^{\infty}{4\cdot(-1)^{n+1}\over n^2+3n+5}$$ converge or diverge?
Solution: $$\lim_{n\to\infty}{4\over n^2+3n+5}=0$$ and $${4\over n^2+3n+5} < {4\over n^2}\to\text{converge}$$ Thus it converges absolutely.