Calculus II 09.05 Alternating Series

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9.5 Alternating Series

• Use the Alternating Series Test to determine whether an infinite series converges.
• Use the Alternating Series Remainder to approximate the sum for an alternating series.
• Classify a convergent series as absolutely or conditionally convergent.
• Rearrange an infinite series to obtain a different sum.

Alternating Series

This section and Section 9.6 describe series with both positive and negative terms. The simplest such series is an alternating series, whose terms alternate in sign. For example, the geometric series

$$ \sum_{n=0}^{\infty} \left (- \frac{1}{2} \right )^{n} $$	$$= \sum_{n=0}^{\infty}(- 1 )^{n} \frac{1}{2^{n}} $$
	$$= 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \cdots$$

is an alternating geometric series with \(r = - \frac{1}{2} \). Alternating series occur in two ways: either the odd terms are negative or the even terms are negative.

Theorem 9.5.1 Alternating Series Test

Let \(a_{n} > 0\). The alternating series

$$ \sum_{n=1}^{\infty} (- 1 )^{n} a_{n}\:\: \text{ and }\:\: \sum_{n=1}^{\infty} (- 1 )^{n+1} a_{n} $$

converge when the two conditions listed below are met. $$1. \: \lim_{n \to \infty} a_{n} = 0 $$ $$2. \: a_{n+1} \leqslant a_{n}\:\: \text{, for all }\:n $$ Proof Consider the alternating series \( \sum (- 1 )^{n+1} a_{n}\). For this series, the partial sum, where \(2n\) is even, is

$$ S_{2n} = (a_{1} - a_{2}) + (a_{3} - a_{4}) + (a_{5} - a_{6}) + \cdots + (a_{2n-1} - a_{2n}) $$

has all nonnegative terms, and therefore \( \{ S_{2n} \} \) is a nondecreasing sequence. But can be rewritten as

$$ S_{2n} = a_{1} - (a_{2} - a_{3}) - (a_{4} - a_{5}) - \cdots - (a_{2n-2} - a_{2n-1}) - a_{2n} $$

which implies that \(S_{2n} \leqslant a_{1}\) for every integer \(n\). Therefore \( \{ S_{2n} \} \) is a bounded, nondecreasing sequence that converges to some value \(L\). Because \(S_{2n-1} - a_{2n} = S_{2n}\) and \(a_{2n} \to 0\), produces

$$ \lim_{n \to \infty} S_{2n-1} $$	$$= \lim_{n \to \infty} S_{2n} + \lim_{n \to \infty} a_{2n} $$
	$$= L + \lim_{n \to \infty} a_{2n} $$
	$$=L.$$

Because both \(S_{2n}\) and \(S_{2n-1}\) converge to the same limit \(L\), it follows that \( \{ S_{2n} \} \) also converges to \(L\). Consequently, the given alternating series converges.

Example 9.5.1 Using the Alternating Series Test

Determine the convergence or divergence for

$$ \sum_{n=1}^{\infty} (- 1 )^{n+1} \frac{1}{n}. $$

Solution The first condition for Theorem 9.5.1 is satisfied because \( \lim a_{n} = \lim 1/n = 0\). The second condition for Theorem 9.5.1 is satisfied because

$$ a_{n+1} = \frac{1}{n+1} \leqslant \frac{1}{n} = a_{n}$$

for all \(n\). The series converges by the Alternating Series Test.

Example 9.5.2 Using the Alternating Series Test with Fractions

Determine the convergence or divergence for

$$ \sum_{n=1}^{\infty} \frac{n}{(- 2 )^{n-1}}. $$

Solution To apply the Alternating Series Test, note that, for \(n \geqslant 1\),

$$ \frac{1}{2} $$	$$ \leqslant \frac{n}{n+1} $$
$$ \frac{2^{n-1}}{2^{n}} $$	$$ \leqslant \frac{n}{n+1} $$
$$ (n+1)2^{n-1} $$	$$ \leqslant n2^{n} $$
$$ \frac{n+1}{2^{n}} $$	$$ \leqslant \frac{n}{2^{n-1}}. $$

Therefore \(a_{n+1} = (n+1)/2^{n} \leqslant n/2^{n-1} = a_{n}\) for all \(n\). Furthermore, by L’Hôpital’s Rule,

$$ \lim_{x \to \infty} \frac{x}{2^{x-1}} = \lim_{x \to \infty} \frac{1}{2^{x-1}(\ln 2)}=0 \rightarrow \lim_{n \to \infty} \frac{n}{2^{n-1}} =0.$$

The series converges by the Alternating Series Test.

Example 9.5.3 When the Alternating Series Test Does Not Apply

a. The alternating series

$$ \sum_{n=1}^{\infty} \frac{(- 1)^{n+1}(n+1)}{n} = \frac{2}{1} - \frac{3}{2} + \frac{4}{3} - \frac{5}{4} + \frac{6}{5}- \cdots $$

passes the second condition for the Alternating Series Test because \(a_{n+1} \leqslant a_{n} \) for all \(n\). However, the series fails the first condition in the Alternating Series Test because the limit does not approach 0. In fact, the series diverges.
b. The alternating series

$$ \frac{2}{1} - \frac{1}{1} + \frac{2}{2} - \frac{1}{2} + \frac{2}{3} - \frac{1}{3} + \frac{2}{4} - \frac{1}{4} + \cdots $$

passes the first condition because \( a_{n}\) approaches 0 as \(n \to \infty \). The series fails the first condition because \(a_{n+1} \nleqslant a_{n} \) for all \(n\). To conclude that the series diverges, you can argue that \(S_{2N}\) equals the \(N\)th partial sum for the divergent harmonic series. This implies that the partial sum sequence diverges. Therefore, the series diverges.

Alternating Series Remainder

For a convergent alternating series, the partial sum \(S_{n} \) can be a useful approximation for the series' sum, \(S\). The error involved in using \(S \approx S_{N}\) is the remainder \(R_{N}=S-S_{N} \).

Theorem 9.5.2 Alternating Series Remainder

If a convergent alternating series satisfies the condition \(a_{n+1} \leqslant a_{n}\), then the absolute value for the remainder \(R_{N}\) involved in approximating the sum \(S\) by \(S_{N} \) is less than, or equal to, the first neglected term. That is,

$$|S-S_{N} |=|R_{N}| \leqslant a_{N+1}. $$

Example 9.5.4 Approximating an Alternating Series' Sum

Approximate the series' sum by the first six terms.

$$ \sum_{n=1}^{\infty} (- 1)^{n+1} \frac{1}{n!} = \frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{5!}- \frac{1}{6!} + \cdots $$

Solution The series converges by the Alternating Series Test because

$$ \frac{1}{(n+1)!} \leqslant \frac{1}{n!} \:\: \text{ and }\:\: \lim_{n \to \infty} \frac{1}{n!} = 0. $$

The first six terms sum to

$$ S_{6} = 1 - \frac{1}{2} + \frac{1}{6} - \frac{1}{24} + \frac{1}{120}- \frac{1}{720} = \frac{91}{144} \approx 0.63194 $$

The answer by the Alternating Series Remainder Theorem is

$$|S-S_{6} |=|R_{6}| \leqslant a_{7} = \frac{1}{5040} \approx 0.0002. $$

The sum \(S\) lies between 0.63194 - 0.0002 and 0.63194 + 0.0002, which yields \(063174 \leqslant S \leqslant 0.63214\).

Example 9.5.5 How many terms are in a series?

Determine how many terms are required to approximate the series' sum with 0.001 error or less.

$$ \sum_{n=1}^{\infty} \frac{(- 1)^{n+1}}{ n^{4}} $$

Solution By the Alternating Series Remainder Theorem

$$|R_{N}| \leqslant a_{N+1} = \frac{1}{(N+1)^{4}}. $$

For an error less than 0.001, \(N\) must satisfy the inequality

$$ \frac{1}{(N+1)^{4}} < 0.001.$$

$$ \frac{1}{(N+1)^{4}} $$	$$ < 0.001 $$
$$ (N+1)^{4} $$	$$ > 1000 $$
$$ N $$	$$ > \sqrt[4]{1000} -1 \approx 4.6$$

Since terms are integers, 0.6 terms can't be selected, 5 will be used. Using 5 terms, the sum is \(S \approx S_{5} \approx 0.94754\), which has less than a 0.001 error.

Absolute and Conditional Convergence

A series may have both positive and negative terms and not be an alternating series. For instance, the series

$$ \sum_{n=1}^{\infty} \frac{\sin n}{ n^{2}} = \frac{\sin 1}{1} + \frac{\sin 2}{4} + \frac{\sin 3}{9}+ \cdots $$

has both positive and negative terms, yet it is not an alternating series. One way to obtain some information about the series' convergence is to investigate it as an absolute value.

$$ \sum_{n=1}^{\infty} \left | \frac{\sin n}{ n^{2}} \right | $$

By direct comparison \( | \sin n| \leqslant 1 \) for all \(n\), therefore

$$ \left | \frac{\sin n}{ n^{2}} \right | \leqslant \frac{1}{n^{2}}, \:\: n \geqslant 1.$$

Therefore, by the Direct Comparison Test, the absolute series converges. Theorem 9.5.3 describe why the original series also converges.

Theorem 9.5.3 Absolute Convergence

If the series \( \sum | a_{n} | \) converges, then the series \(\sum a_{n}\) also converges.
Proof Because \( 0 \leqslant a_{n} + | a_{n} | \leqslant 2| a_{n} | \) for all \(n\), the series

$$ \sum_{n=1}^{\infty} ( a_{n} + | a_{n} |) $$

converges by comparison with the convergent series

$$ \sum_{n=1}^{\infty} 2| a_{n} | .$$

Furthermore, because \( a_{n} = ( a_{n} + |a_{n}|)-|a_{n}| \), the series can be written as

$$ \sum_{n=1}^{\infty} a_{n} = \sum_{n=1}^{\infty} ( a_{n} + |a_{n}|)- \sum_{n=1}^{\infty} |a_{n}| $$

where both series on the right converge. Therefore, it follows that the series converges.

The converse is not true. For instance, the alternating harmonic series

$$ \sum_{n=1}^{\infty} \frac{(- 1)^{n+1}}{n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots $$

converges by the Alternating Series Test. Yet the harmonic series diverges. This convergence type is called conditional.

Definition 9.5.1 Absolute and Conditional Convergence

1. The series \( \sum a_{n}\) is absolutely convergent when \( \sum |a_{n}| \) converges.
2. The series \( \sum a_{n}\) is conditionally convergent when \( \sum a_{n} \) converges but \( \sum |a_{n}| \) diverges.

Example 9.5.6 Absolute and Conditional Convergence with Factorials

Determine whether each series is convergent or divergent. Classify any convergent series as absolutely or conditionally convergent. $$a. \:\: \sum_{n=1}^{\infty} \frac{(- 1 )^{n} n!}{ 2^{n}} = \frac{0!}{2^{0}} - \frac{1!}{2^{1}} + \frac{2!}{2^{2}} - \frac{3!}{2^{3}} + \cdots$$ $$b. \:\: \sum_{n=1}^{\infty} \frac{(- 1 )^{n}}{ \sqrt{n}} = - \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} - \cdots$$ Solution
a. This is an alternating series, but the Alternating Series Test does not apply because the limit for the \(n\)th term is not zero. The series diverges by the \(n\)-Term Test for Divergence.
b. This series is convergent by the Alternating Series Test. Moreover, because the \(p\)-series

$$ \sum_{n=1}^{\infty} \left | \frac{(- 1 )^{n}}{ \sqrt{n}} \right | = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{4}} + \cdots$$

diverges and is conditionally convergent.

Example 9.5.7 Absolute and Conditional Convergence with Natural Logs

Determine if the series is convergent or divergent. Classify any convergent series as absolutely or conditionally convergent. $$a. \:\: \sum_{n=1}^{\infty} \frac{(- 1 )^{n(n+1)/2}}{ 3^{n}} = - \frac{1}{3} - \frac{1}{9} + \frac{1}{27} + \frac{1}{81} - \cdots$$ $$b. \:\: \sum_{n=1}^{\infty} \frac{(- 1 )^{n}}{ \ln(n+1)} = - \frac{1}{\ln 2} + \frac{1}{\ln 3} - \frac{1}{\ln 4} + \frac{1}{\ln 5} - \cdots$$ Solution
a. The (a) series is not an alternating series because the signs change in pairs. Note that

$$ \sum_{n=1}^{\infty} \left | \frac{(- 1 )^{n(n+1)/2}}{ 3^{n}} \right | = \sum_{n=1}^{\infty} \frac{1}{ 3^{n}} $$

is a convergent geometric series, with

$$r=\frac{1}{3}. $$

By the Absolute Convergence Theorem the series is absolutely convergent, and therefore convergent for this example.
b. The series converges by The Alternating Series Test. The absolute series

$$\sum_{n=1}^{\infty} \left | \frac{(- 1 )^{n}}{ \ln(n+1)} \right | = \frac{1}{\ln 2} + \frac{1}{\ln 3} + \frac{1}{\ln 4}+ \cdots $$

diverges when terms are compared with a harmonic series. Therefore, the series is conditionally convergent.

Rearranging a Series

A finite sum such as

$$1+3-2+5-4 $$

can be rearranged without changing the sum's value. This is not necessarily true for an infinite series –– it depends on whether the series is absolutely convergent or conditionally convergent.
1. If a series is absolutely convergent, then its terms can be rearranged in any order without changing the sum for the series.
2. If a series is conditionally convergent, then its terms can be rearranged to give a different sum.
The second case is described in Example 9.5.8.

Example 9.5.8 Rearranging a Series

The alternating harmonic series converges to \(\ln 2\). As shown,

$$ \sum_{n=1}^{\infty} (- 1 )^{n+1} \frac{1}{n} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2.$$

Rearrange the series to produce a different sum.
Solution Consider the rearrangement below.

$$S_{n} $$	$$= 1- \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} -\frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \frac{1}{7} - \frac{1}{14} - \cdots $$
	$$= \left ( 1- \frac{1}{2} \right ) - \frac{1}{4} + \left ( \frac{1}{3} - \frac{1}{6}\right ) -\frac{1}{8} + \left ( \frac{1}{5} - \frac{1}{10} \right ) - \frac{1}{12} + \left ( \frac{1}{7} - \frac{1}{14} \right ) - \cdots$$
	$$= \frac{1}{2} - \frac{1}{4} + \frac{1}{6} -\frac{1}{8} + \frac{1}{10} - \frac{1}{12} + \frac{1}{14} - \cdots$$
	$$= \frac{1}{2} \left ( 1 - \frac{1}{2} + \frac{1}{3}- \frac{1}{4} + \frac{1}{5} - \frac{1}{6} +\frac{1}{7} - \cdots \right )$$
	$$= \frac{1}{2} (\ln 2) $$

By rearranging the terms, you obtain a sum that is half the original sum.

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Parent Article: Calculus II 09 Infinite Series