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Wize University Calculus 2 Textbook > Sequences and Series

Absolute Convergence & Alternating Series Test

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Absolute Convergence Test

Note: If ∑n=1∞∣an∣\sum\limits^\infty_{n=1}|a_n| converges, then ∑n=1∞an\sum\limits^\infty_{n=1}a_n converges.

Wize Concept

Definitions:

  • We say that a series, ∑n=1∞an\sum\limits^\infty_{n=1}a_n is absolutely convergent if the series ∑n=1∞∣an∣\sum\limits^\infty_{n=1}|a_n| converges
  • We say that a series, ∑n=1∞an\sum\limits^\infty_{n=1}a_n is conditionally convergent if it converges but is not absolutely convergent
*If the terms in the series are all positive, we don't have to make this distinction.

Absolute Convergence Test

If it is easier to work with ∣an∣\left|a_n\right|, then determine whether ∑n=1∞∣an∣\sum\limits^\infty_{n=1}|a_n| converges.
If it does converge, then the original series ∑n=1∞an\sum\limits^\infty_{n=1}a_n must also converge.

Identifying Clues

  • If the series involves negative and positive terms
  • The question will ask specifically for "absolute convergence", "conditional convergence", or divergence

Example: Absolute Convergence Test

Determine whether the following series are absolutely convergent, conditionally convergent, or divergent:

a.) ∑n=1∞(−1)nn\sum\limits^\infty_{n=1}\frac{(-1)^n}{n}


b.) ∑n=0∞(−2)n\sum\limits^\infty_{n=0}(-2)^n



c.) ∑n=1∞(−1)nn53\sum\limits^\infty_{n=1}\frac{(-1)^n}{\sqrt[3]{n^5}}

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Alternating Series Test

Consider a series with terms that have alternating signs ∑n=1∞an=∑n=1∞(−1)n+1bn,\sum\limits^\infty_{n=1}a_n=\sum\limits^\infty_{n=1}(-1)^{n+1}b_n,
where the terms bnb_n are positive.
  • If bn+1≤bnb_{n+1}\le b_n (the terms do not increase) after a certain point and
  • If lim⁡n→∞bn=0\lim\limits_{n\rightarrow\infty}b_n =0
Then the series converges.


Identifying Clues:

  • The terms have alternating signs

Practice: AST

Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?

Extra Practice
Practice: Alternating Series Test
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice Problem
Determine whether the following series are absolutely convergent, conditionally convergent, or divergent:
a.) ∑n=1∞(−1)nn\sum\limits^\infty_{n=1}\frac{(-1)^n}{n}
b.) ∑n=0∞(−2)n\sum\limits^\infty_{n=0}(-2)^n
Practice: Alternating Series Test
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice Problem
Example: Absolute Convergence Test
Determine whether the following series are absolutely convergent, conditionally convergent, or divergent:
a.) ∑n=1∞(−1)nn\sum\limits^\infty_{n=1}\frac{(-1)^n}{n}
Practice: Alternating Series Test
Practice: AST
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice: Alternating Series Test
Practice: AST
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice: Alternating Series Test
Practice: AST
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice: Alternating Series Test
Practice: AST
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice Problem
Example: Absolute Convergence Test
Determine whether the following series are absolutely convergent, conditionally convergent, or divergent:
a.) ∑n=1∞(−1)nn\sum\limits^\infty_{n=1}\frac{(-1)^n}{n}
Practice: Alternating Series Test
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent? If convergent, determine whether it converges conditionally or absolutely.
Practice: Alternating Series Test
Practice: AST
Is the series∑n=2∞(−1)n+1n2n3+3\sum\limits^\infty_{n=2}(-1)^{n+1}\frac{n^2}{n^3+3} convergent or divergent?
Practice Problem
Example: Absolute Convergence Test
Determine whether the following series are absolutely convergent, conditionally convergent, or divergent:
a.) ∑n=1∞(−1)nn\sum\limits^\infty_{n=1}\frac{(-1)^n}{n}