Alternating Series

Calculus 2

Alternating Series

An alternating series is a series whose terms alternate between negative and positive (or non-negative and non-positive) values. The nnth term of an alternating series is of the form:

an=(1)n+1bn   or    an=(1)nbna_n = (-1)^{n+1}b_n \ \ \text{ or } \ \ \ a_n=(-1)^nb_n

where bn=anb_n=|a_n| is a positive number.

When working with series whose terms alternate, we are interested in isolating the alternating factor, or the portion of the general term that produces the alternating effect.

Commonly, the alternating factor is (1)n(-1)^n. However, the alternating factor may be disguised. Take a look at the following terms:

(1)n1=(1)n(1)n+2=(1)n(4)n=(1)n4ncos(πn)=(1)n(2)n+1=(1)n2n+1\begin{aligned} (-1)^{n-1} &= -(-1)^n \\ (-1)^{n+2} &= (-1)^n \\ (-4)^{n} &= (-1)^n \cdot 4^n \\ \cos(\pi n) &= (-1)^n \\ (-2)^{n+1} &= -(-1)^n \cdot 2^{n+1} \\ \end{aligned}

Alternating Series Test (AST)

Try to visualize the following three scenarios (as well as the absolute value of each):

  1. A series whose terms alternate to zero.
  2. A series whose terms alternate to a non-zero finite value.
  3. A series whose terms alternate to an infinite value.

These scenarios are what leads us to the Alternating Series Test (AST), which states that the series

n=1(1)n+1bn=b1b2+b3b4+   (where bn>0)\sum_{n=1}^\infty (-1)^{n+1}b_n = b_1 - b_2 + b_3 - b_4 + \cdots \ \ \ \text{(where } b_n > 0 \text{)}

converges if the following two conditions are satisfied:

  1. bnbn+1b_n \geq b_{n+1} for all nNn \geq N, for some integer NN.
  2. limnbn=0\lim_{n \to \infty} b_n = 0.

In simple terms, when the alternating factor is removed, are the terms in the series positive and decreasing toward zero? If so, then the series is convergent according to the AST.

Note that the AST cannot determine divergence, although in many divergent cases, the nnth Term Test will be useful.

Example 1

Determine whether the series n=2(1)nn2\sum_{n=2}^\infty \frac{(-1)^n}{n^2} converges or diverges using the Alternating Series Test.

View Answer

Converges

View Solution

The series satisfies the AST:

  • bn=1n2b_n = \frac{1}{n^2} is positive and decreasing.
  • limnbn=0\lim_{n \to \infty} b_n = 0.

Thus, the series converges.

Example 2

Determine whether the series n=3(1)n+1n\sum_{n=3}^\infty \frac{(-1)^{n+1}}{n} converges or diverges using the Alternating Series Test.

View Answer

Converges

View Solution

The series satisfies the AST:

  • bn=1nb_n = \frac{1}{n} is positive and decreasing.
  • limnbn=0\lim_{n \to \infty} b_n = 0.

Thus, the series converges.

Example 3

Determine whether the series n=1cos(πn)n+3\sum_{n=1}^\infty \frac{\cos(\pi n)}{n+3} converges or diverges using the Alternating Series Test.

View Answer

Converges

Example 4

Determine whether the series n=1(1)n+1n3\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^3} converges or diverges using the Alternating Series Test.

View Answer

Converges

Example 5

Determine whether the series n=1(2)n+1n5\sum_{n=1}^\infty \frac{(-2)^{n+1}}{n^5} converges or diverges.

View Answer

Inconclusive by the Alternating Series Test; Divergent by nnth Term Test

Example 6

Determine whether the series n=1(1)nn2n\sum_{n=1}^\infty \frac{(-1)^n \cdot n}{2^n} converges or diverges.

View Answer

Converges by the Alternating Series Test


Absolute and Conditional Convergence

As we've seen, when the terms of a series are positive and negative, the series may or may not converge. For example:

  • The geometric series n=05(14)n\sum_{n=0}^\infty 5 \left( -\frac{1}{4} \right)^n converges.
  • The geometric series n=0(54)n\sum_{n=0}^\infty \left( -\frac{5}{4} \right)^n diverges.

If we ignore the alternating factor from both series, we see a pattern arise:

If n=0an converges, then n=0an converges.\text{If } \sum_{n=0}^\infty |a_n| \text{ converges, then } \sum_{n=0}^\infty a_n \text{ converges.}

A series an\sum a_n is said to converge absolutely (or is absolutely convergent) if the corresponding series of absolute values an\sum |a_n| converges. Thus, if a series is absolutely convergent, it must also be convergent.

We call a series conditionally convergent if an\sum a_n converges but an\sum |a_n| diverges.

One interesting thing about conditionally convergent series (which we will not explore here) is that they will have different sums if you rearrange the order of the terms of the series.

Example 7

Determine whether the series n=1(1)nn\sum_{n=1}^\infty \frac{(-1)^n}{n} converges or diverges. If it converges, determine the type.

View Answer

Conditionally convergent

View Solution

The alternating series n=1(1)nn\sum_{n=1}^\infty \frac{(-1)^n}{n} converges by the AST. However, an=1n\sum |a_n| = \sum \frac{1}{n} diverges (harmonic series). Thus, the series is conditionally convergent.

Example 8

Determine whether the series n=1(1)nn2\sum_{n=1}^\infty \frac{(-1)^n}{n^2} converges or diverges. If it converges, determine the type.

View Answer

Absolutely convergent

View Solution

The series converges by the AST. Additionally, an=1n2\sum |a_n| = \sum \frac{1}{n^2} converges (p-Series Test, p=2>1p = 2 > 1). Thus, the series is absolutely convergent.

Example 9

Determine whether the series n=1(3)nn2\sum_{n=1}^\infty \frac{(-3)^n}{n^2} converges or diverges. If it converges, determine the type.

View Answer

Diverges

Example 10

Determine whether the series n=1cos(πn)n\sum_{n=1}^\infty \frac{\cos(\pi n)}{\sqrt{n}} converges or diverges. If it converges, determine the type.

View Answer

Conditionally convergent


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