Mooculus calculus-2: Sequence and progression study Note 4. Alternating series

This course (MOOCULUS-2 "sequences and Series") was taught by Ohio State University on the Coursera platform in 2014.

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Summary

• If $$\sum_{n=1}^\infty |a_n|$$ converges (i.e. absolutelyconvergent), then $$\sum_{n=1}^\infty a_n$$ converges (i.e. conditionally convergent).
• Suppose that $ (a_n) $ was a decreasing sequence of positive numbers and $$\lim_{n\to\infty}a_n=0$$ then th E Alternating series $$\sum_{n=1}^\infty ( -1) ^{n+1} a_n$$ converges.
• For a 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 $${\ln n\o Ver n} > {1\over n}\to\text{diverge}$$ 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 $${\l n N\over n^3} < {n\over n^3}={1\over n^2}\to\text{converge}$$ 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 conver Ges.

5. Approximate $$\sum_{n=1}^\infty ( -1) ^{n+1}{1\over n^3}$$ to the 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 the 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, 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 "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.

Mooculus calculus-2: Sequence and progression study Note 4. Alternating series