Section 6.4 Alternating Series
¶Next we consider series with both positive and negative terms, but in a regular pattern: they alternate signs. For example:
\(\ds\sum_{n=1}^{\infty} (-1)^{n-1} \frac{2}{n+1} = 1 - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \dots\)
\(\ds\sum_{n=1}^{\infty} (-1)^n \frac{2}{n+1} = -1 + \frac{2}{3} - \frac{2}{4} + \frac{2}{5} - \dots\)
\(\ds\sum_{n=0}^{\infty} (-1)^n 2^n =1 - 2 + 4 - 8 + \dots\)
\(\ds\sum_{n=4}^{\infty} (-1)^{n-1}\frac{n}{n+2} = -\frac{4}{6} + \frac{5}{7} - \frac{6}{8} + \frac{7}{9} - \dots\)
Definition 6.45. Alternating Series.
An alternating series has the form
\begin{equation*}\sum (-1)^n a_n\end{equation*}
where \(a_n\) are all positive and the first index is arbitrary.
A well-known example of an alternating series is the alternating harmonic series:
Definition 6.46. Alternating Harmonic Series.
A series of the form
\begin{equation*}\sum_{n=1}^\infty {\frac{(-1)^{n-1}}{n}}= 1 - \frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{(-1)^{n-1}}{n} + \dots\end{equation*}
is called an alternating harmonic series.
In the alternating harmonic series the magnitude of the terms decrease, that is, \(\ds |a_n|\) forms a decreasing sequence, although this is not required in an alternating series. Recall that for a series with positive terms, if the limit of the terms is not zero, the series cannot converge; but even if the limit of the terms is zero, the series still may not converge. It turns out that for alternating series, the series converges exactly when the limit of the terms is zero. In Figure6.4, we illustrate what happens to the partial sums of the alternating harmonic series. Because the sizes of the terms \(\ds a_n\) are decreasing, the odd partial sums \(\ds s_1\text{,}\) \(\ds s_3\text{,}\) \(\ds s_5\text{,}\) and so on, form a decreasing sequence that is bounded below by \(\ds s_2\text{,}\) so this sequence must converge. Likewise, the even partial sums \(\ds s_2\text{,}\) \(\ds s_4\text{,}\) \(\ds s_6\text{,}\) and so on, form an increasing sequence that is bounded above by \(\ds s_1\text{,}\) so this sequence also converges. Since all the even numbered partial sums are less than all the odd numbered ones, and since the “jumps” (that is, the \(\ds a_i\) terms) are getting smaller and smaller, the two sequences must converge to the same value, meaning the entire sequence of partial sums \(\ds s_1,s_2,s_3,\ldots\) converges as well.
The same argument works for any alternating sequence with terms that decrease in absolute value. The Alternating Series Test is worth calling a theorem.
Theorem 6.47. Alternating Series Test.
Suppose that \(\ds\{a_n\}_{n=1}^\infty\) is a decreasing sequence of positive numbers and \(\ds\lim_{n\to\infty}a_n=0\text{.}\) Then the alternating series \(\ds\sum_{n=1}^\infty (-1)^{n-1} a_n\) converges.
Proof.
The odd-numbered partial sums, \(\ds s_1, s_3, s_5,\ldots, s_{2k+1},\ldots\text{,}\) form a decreasing sequence, because \(\ds s_{2k+3}=s_{2k+1}-a_{2k+2}+a_{2k+3}\le s_{2k+1}\text{,}\) since \(\ds a_{2k+2}\ge a_{2k+3}\text{.}\) This sequence is bounded below by \(\ds s_2\text{,}\) so it must converge, to some value \(L\text{.}\) Likewise, the partial sums \(\ds s_2, s_4, s_6,\ldots,s_{2k},\ldots\text{,}\) form an increasing sequence that is bounded above by \(\ds s_1\text{,}\) so this sequence also converges, to some value \(M\text{.}\) Since \(\ds\lim_{n\to\infty} a_n=0\) and \(\ds s_{2k+1}= s_{2k}+a_{2k+1}\text{,}\)
\begin{equation*}L=\lim_{k\to\infty}s_{2k+1}=\lim_{k\to\infty}(s_{2k}+a_{2k+1})= \lim_{k\to\infty}s_{2k}+\lim_{k\to\infty}a_{2k+1}=M+0=M\text{,}\end{equation*}
so \(L=M\text{;}\) the two sequences of partial sums converge to the same limit, and this means the entire sequence of partial sums also converges to \(L\text{.}\)
Another useful fact is implicit in this discussion. Suppose that
\begin{equation*}L=\sum_{n=1}^\infty (-1)^{n-1} a_n\end{equation*}
and that we approximate \(L\) by a finite part of this sum, say
\begin{equation*}L\approx \sum_{n=1}^N (-1)^{n-1} a_n\text{.}\end{equation*}
Because the terms are decreasing in size, we know that the true value of \(L\) must be between this approximation and the next one, that is, between
\begin{equation*}\sum_{n=1}^N (-1)^{n-1} a_n \hbox{and} \sum_{n=1}^{N+1} (-1)^{n-1} a_n\text{.}\end{equation*}
Depending on whether \(N\) is odd or even, the second will be smaller or larger than the first.
Example 6.48. Approximating a Series.
Approximate the sum of the alternating harmonic series to within 0.05.
We need to go to the point at which the next term to be added or subtracted is \(1/10\text{.}\) Adding up the first nine and the first ten terms we get approximately \(0.746\) and \(0.646\text{.}\) These are \(1/10\) apart, so the value halfway between them, 0.696, is within 0.05 of the correct value.
Note: We have considered alternating series with first index 1, and in which the first term is positive, but a little thought shows this is not crucial. The same test applies to any similar series, such as \(\ds\sum_{n=0}^\infty (-1)^n a_n\text{,}\) \(\ds\sum_{n=1}^\infty (-1)^n a_n\text{,}\) \(\ds\sum_{n=17}^\infty (-1)^n a_n\text{,}\) etc.
Exercises for Section6.4.
Exercise 6.4.1.
Determine whether the following series converge or diverge.
\(\ds\sum_{n=1}^\infty {(-1)^{n-1}\over 2n+5}\)
AnswerSolutionconverges
Let \(a_n = \dfrac{1}{2n+5}\text{.}\) Then
\begin{equation*}\lim_{n\to\infty}a_n = \lim_{n\to\infty} \frac{1}{2n+5} = 0\text{,}\end{equation*}
and we also see that the sequence \(\{a_n\}_{n=1}^{\infty}\) is non-increasing:
\begin{equation*}\begin{split} a_n \amp \geq a_{n+1} \\ \frac{1}{2n+5} \amp \geq \frac{1}{2(n+1)+5} = \frac{1}{2n+7} \\ 2n+ 7 \amp \geq 2n + 5 \\ 7 \amp \geq 5, \end{split}\end{equation*}
which is true. Therefore, by the Alternating Series Test, the series \(\displaystyle \sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{2n+5}\) converges.
\(\ds\sum_{n=4}^\infty {(-1)^{n-1}\over \sqrt{n-3}}\)
AnswerSolutionconverges
Let \(a_n = \dfrac{1}{\sqrt{n-3}}\text{.}\) We first calculate
\begin{equation*}\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{\sqrt{n-3}} = 0\text{.}\end{equation*}
Next, we check whether or not the sequence \(\{a_n\}_{n=4}^{\infty}\) is decreasing:
\begin{equation*}\begin{split} a_n \amp \geq a_{n+1} \\ \frac{1}{\sqrt{n-3}} \amp \geq \frac{1}{\sqrt{n-2}} \\ \sqrt{n-2} \amp \geq \sqrt{n-3} \\ -2 \amp \geq -3, \end{split}\end{equation*}
which is true. Both conditions of the Alternating Series Test are satisfied, and so the series \(\displaystyle \sum_{n=4}^{\infty} \dfrac{(-1)^{n-1}}{\sqrt{n-3}}\) converges.
\(\ds\sum_{n=1}^\infty (-1)^{n-1}{n\over 3n-2}\)
AnswerSolutiondiverges
We now take \(a_n = \dfrac{n}{3n-2}\text{,}\) and first calculate the limit:
\begin{equation*}\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{n}{3n-2} = \lim_{n\to\infty} \frac{1}{3-\frac{2}{n}} = \frac{1}{3} \neq 0\text{.}\end{equation*}
Therefore, by the \(n\)-th Term Test, the series diverges.
\(\ds\sum_{n=1}^\infty (-1)^{n-1}{\ln n\over n}\)
AnswerSolutionconverges
Let \(a_n = \dfrac{\ln(n)}{n}\text{.}\) We first compute
\begin{equation*}\lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln(n)}{n} \Heq \lim_{n\to\infty} \frac{1}{n} \cdot \frac{1}{1} = 0.\end{equation*}
We now check whether or not \(\left\{a_n\right\}\) is a decreasing sequence. Since
\begin{equation*}f(x) = \frac{\ln(x)}{x} \implies f'(x) = \frac{1-\ln(x)}{x^2}\end{equation*}
is decreasing on \((e, \infty)\text{,}\) this means that the sequence is decreasing for \(n \ge e\text{.}\) Hence, by the Alternating Series Test, the series converges.
Exercise 6.4.2.
Approximate each series to within the given error.
\(\ds\sum_{n=1}^\infty (-1)^{n-1}{1\over n^3}\text{,}\) 0.005
AnswerSolution\(0.904\)
We first notice that the series converges by Alternating Series Test. We now calculate the following partial sums:
\begin{equation*}\begin{array}{l|lllllll} n \amp 1 \amp 2 \amp 3 \amp 4 \amp 5 \amp 6 \amp \dots \\ \hline s_n \amp 1 \amp 0.875 \amp 0.912...\amp 0.896... \amp 0.904... \amp 0.899... \amp \dots \end{array}\end{equation*}
We have that \(|s_6-s_5| = |a_6| = 0.004629... \lt 0.005\text{.}\) Therefore, \(s_5 \approx 0.904\) is within 0.005 of the true sum.
\(\ds\sum_{n=1}^\infty (-1)^{n-1}{1\over n^4}\text{,}\) 0.005
AnswerSolution\(0.946\)
We first notice that the series converges by Alternating Series Test. We now calculate the following partial sums:
\begin{equation*}\begin{array}{l|lllllll}n \amp 1 \amp 2\amp 3\amp 4 \amp \dots \\ \hlines_n \amp 1 \amp 0.9375\amp 0.94584..\amp 0.945939...\dots\end{array}\end{equation*}
We have that \(|s_4-s_3| = |a_4| = 0.0039... \le 0.005 \text{.}\) Therefore, \(s_3 \approx 0.946\) is within 0.005 of the true sum.
