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Summary
- If \sum_{n=1}^\infty |a_n|converges (i.e. absolutely convergent), then \sum_{n=1}^\infty a_nconverges (i.e. conditionally convergent).
- Suppose that (a_n) is a decreasing sequence of positive numbers and \lim_{n\to\infty}a_n=0Then the alternating series \sum_{n=1}^\infty (-1)^{n+1} a_nconverges.
- For an alternating series s_n=\sum_{n=1}^{\infty}(-1)^n\cdot a_nthe test steps:
- If \lim_{n\to\infty}a_n\neq0then it diverges;
- If \lim_{n\to\infty}a_n=0and a_n converges, then it absolutely converges;
- If \lim_{n\to\infty}a_n=0and a_n diverges, then it conditionally converges.
- If \lim_{n\to\infty}a_n\neq0
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.
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