10.5: Alternating Series

Up to this point, we have focused on series whose terms were all positive. For example, in Section 10.3 we learned the Integral Test, which applies to a series with positive terms. Then in Section 10.4, we introduced a pair of tests that permits us to compare two series whose terms are all positive. But what if some terms are negative? In this section we define alternating series, introduce a test to establish whether they converge, and learn to bound the errors in their partial sums. We discuss the following topics:

Alternating Series
Alternating Series Error Bound

Alternating Series

Alternating series are series whose terms alternate in sign. Consider the following examples of alternating series: \begin{align} -1 + 2 - 3 + 4 - 5 + \cdots &= \sum_{n = 1}^\infty (-1)^n \, n \cma \label{eq:alt-n} \nl 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots &= \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \pd \label{eq:alt-harmonic} \end{align} In general, an alternating series takes the form \[\sum_{n = 1}^\infty (-1)^{n + 1} \, b_n \or \sum_{n = 1}^\infty (-1)^n \, b_n \cma\] where \(b_n \gt 0.\) We call the term \((-1)^n\) or \((-1)^{n + 1}\) the alternator because it changes sign depending on whether \(n\) is odd or even. \(\eqrefer{eq:alt-harmonic}\) is the Alternating Harmonic series since it's the sum of reciprocals whose signs are alternating.

Let's now discuss a test to determine whether an alternating series converges. The Alternating Series Test asserts that \(\sum_{n = 1}^\infty (-1)^n \, b_n\) (where \(b_n\) is positive) converges if \(\lim_{n \to \infty} b_n = 0\) and \(\{b_n\}\) is a decreasing sequence (that is, \(b_{n + 1} \leq b_n\)). In words, the test states that an alternating series converges if its terms are shrinking in magnitude to \(0.\)

ALTERNATING SERIES TEST

The Alternating Series Test states that \[\sum_{n = 1}^\infty (-1)^n \, b_n \cmaa b_n \gt 0 \cma\] converges if \(\lim_{n \to \infty} b_n = 0\) and \(b_{n + 1} \leq b_n\) for all \(n \geq N\) for some nonnegative integer \(N.\)

The Alternating Series Test can only establish convergence; it cannot prove that a series diverges. If a series does not meet the conditions of the Alternating Series Test, then the test is inconclusive. In many situations, if \(\lim_{n \to \infty} b_n \ne 0,\) then we can use the Divergence Test to prove that \(\sum (-1)^n \, b_n\) diverges.

Figure 1 enables us to grasp the intuition behind the Alternating Series Test by using a number line. Suppose that the infinite series \(\sum_{n = 1}^\infty (-1)^{n + 1} \, b_n\) (where \(b_n\) is positive) converges to \(S.\) Let \(S_N\) denote the \(N\)th partial sum of this alternating series. First, we add the term \(b_1\) to \(0;\) doing so, we arrive at the value \(S_1\) on the number line. Then we subtract the term \(b_2,\) so \(S_2\) is to the left of \(S_1.\) Adding \(b_3,\) we attain the value \(S_3\) to the right of \(S_2.\) This pattern repeats—we continue adding or subtracting incrementally smaller numbers since \(b_n\) is decreasing and \(b_n \to 0.\) In doing so, the value of \(S_N\) gets closer and closer to the number \(S.\) Accordingly, if we repeat this process infinitely many times—that is, letting \(N \to \infty\)—then it appears that \(S_N \to S.\) We observe the following: For odd \(N,\) \(S_N\) always lies to the right of \(S.\) Note that the partial sums \(S_1, S_3,\) \(S_5, \dots\) are decreasing toward \(S.\) Conversely, for even \(N,\) \(S_N\) always lies to the left of \(S.\) And the partial sums \(S_2, S_4,\) \(S_6, \dots\) are increasing toward \(S.\) Hence, the odd partial sums and even partial sums both seem to converge to \(S.\) We use this observation to motivate the following proof.

PROOF Let's split the proof into two cases—one for even \(N\) and one for odd \(N.\) Since \(b_n\) is decreasing, we have \(b_2 \geq b_3 \geq b_4\) \(\geq b_5 \geq \dots \geq b_{2N}\) and so \[ \ba S_2 &= b_1 - b_2 \geq 0 \cma \nl S_4 &= S_2 + (b_3 - b_4) \geq S_2 \cma \nl & \vdotss \nl S_{2N} &= S_{2N - 2} + (b_{2N - 1} - b_{2N}) \geq S_{2N - 2} \pd \ea \] But observe that \[S_{2N} = b_1 - (b_2 - b_3) - (b_4 - b_5) - \cdots - (b_{2N - 2} - b_{2N - 1}) - b_{2N} \pd\] In each set of parentheses, the difference is positive (or \(0\)). Accordingly, \(0 \leq S_{2N} \leq b_1.\) Because \[S_{2N} = \sum_{i = 1}^{2N} (-1)^{i + 1} \, b_i \cma\] we see \[\lim_{N \to \infty} S_{2N} = \lim_{N \to \infty} \sum_{i = 1}^{2N} (-1)^{i + 1} \, b_i = \sum_{n = 1}^\infty (-1)^{n + 1} \, b_n \leq b_1 \pd\] Since \(S_{2N}\) is increasing (and therefore monotonic) and bounded, it converges by the Monotonic Sequence Theorem. Now let \(S = \lim_{N \to \infty} S_{2N}.\) For odd \(N,\) we see \[ \ba \lim_{N \to \infty} S_{2N + 1} &= \lim_{N \to \infty} \par{S_{2N} + b_{2N + 1}} \nl &= \lim_{N \to \infty} S_{2N} + \lim_{N \to \infty} b_{2N + 1} \nl &= S + 0 = S \cma \ea \] where the last step is true because we require \(\lim_{n \to \infty} b_n = 0.\) Hence, for both even and odd \(N,\) the alternating series \(\sum_{n = 1}^\infty (-1)^{n + 1} \, b_n\) converges. \[\qedproof\]

EXAMPLE 1

Test the following series for convergence or divergence: \[\sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{\sqrt{n}} \pd\]

We identify \(b_n = 1/\sqrt{n},\) and the alternator is \((-1)^{n + 1}.\) For \(n \geq 1\) we see \[\underbrace{\frac{1}{\sqrt{n + 1}}}_{b_{n + 1}} \leq \underbrace{\frac{1}{\sqrt{n}}}_{b_n}. \] Therefore, \(b_n\) is decreasing for all \(n \geq 1.\) Additionally, \(\lim_{n \to \infty } (1/\sqrt{n} \, ) = 0,\) so by the Alternating Series Test, \[\[\sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{\sqrt{n}} \pd\] \convBoxed\]

It turns out that the Alternating Harmonic series, as in \(\eqref{eq:alt-harmonic},\) satisfies \[\lim_{n \to \infty} \frac{1}{n} = 0 \and \frac{1}{n + 1} \leq \frac{1}{n} \pd\] In words, the terms of the Alternating Harmonic series are shrinking in magnitude toward \(0.\) Thus, the Alternating Test asserts that the series converges—unlike the standard Harmonic series, which diverges.

EXAMPLE 2

Test the following series for convergence or divergence: \[\sum_{n = 1}^\infty (-1)^{n + 1} \frac{\sqrt{n}}{n + 2} \pd\]

We identify \(b_n = \sqrt{n}/(n + 2),\) but it isn't clear whether this function is decreasing. Let's therefore compare \(b_n\) to the function \(f(x) = \sqrt x/(x + 2),\) which is continuous for all nonnegative \(x.\) Differentiating shows \[ f'(x) = \frac{\dfrac{x + 2}{2 \sqrt x} - \sqrt x}{(x + 2)^2} = \frac{-x + 2}{2 (x + 2)^2 \sqrt x} \cma \] which is negative for all \(x \gt 2.\) Hence, \(f\) is decreasing on \([2, \infty),\) meaning \(b_{n + 1} \leq b_n\) for all \(n \geq 2.\) Since \(\lim_{n \to \infty} b_n = 0,\) the Alternating Series Test states that \[\sum_{n = 1}^\infty (-1)^{n + 1} \frac{\sqrt{n}}{n + 2} \convBoxed\]

REMARK Note that \(b_n\) is decreasing for \(n \geq 2,\) but the series' index begins at \(n = 1.\) Can we still use the Alternating Series Test? Yes; we only require that the terms of \(\{b_n\}\) eventually be decreasing to \(0.\) Writing out some terms shows \[\sum_{n = 1}^\infty (-1)^{n + 1} \frac{\sqrt{n}}{n + 2} = \underbrace{\frac{1}{3}}_{0.333 \dots} - \underbrace{\frac{\sqrt 2 }{4}}_{0.354 \dots} + \underbrace{\frac{\sqrt 3 }{5}}_{0.346 \dots} - \underbrace{\frac{1}{3}}_{0.333\dots} + \underbrace{\frac{\sqrt 5}{7}}_{0.319 \dots} + \cdots \pd \] After the second term \(-\sqrt 2/4\), each incremental term is decreasing in magnitude to \(0.\) Because \(\sum_{n = 2}^\infty (-1)^{n + 1} \, b_n\) converges, \(\sum_{n = 1}^\infty (-1)^{n + 1} \, b_n\) also converges. (Adding or subtracting a finite number of numbers to a series doesn't alter whether it converges or diverges.)

EXAMPLE 3

Test the following series for convergence or divergence: \[\sum_{n = 1}^\infty \cos(\pi n) \frac{n}{n^2 + 1} \pd\]

Note that \(\cos(\pi n) \) is the alternator in this series because \(\cos(\pi n)\) \(= (-1)^n.\) We identify \(b_n = n/(n^2 + 1).\) To determine whether the terms of the sequence \(\{b_n\}\) are decreasing, we compare \(b_n\) to the continuous function \(f(x) = x/(x^2 + 1).\) Differentiating \(f(x),\) we get \[ f'(x) = \frac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} \nl = \frac{1 - x^2}{(x^2 + 1)^2} \cma \] which is negative for all \(x \gt 1.\) Therefore, \(f\) is decreasing over \([1, \infty),\) so \(b_{n + 1} \leq b_n\) for all \(n \geq 1.\) And since \(\lim_{n \to \infty} b_n = 0,\) the Alternating Series Test states that \[\sum_{n = 1}^\infty \cos(\pi n) \frac{n}{n^2 + 1} \convBoxed\]

EXAMPLE 4

Test the following series for convergence or divergence: \[\sum_{n = 1}^\infty (-1)^n \frac{n^3}{3n^3 + 1} \pd\]

We see \(b_n = n^3/\par{3n^3 + 1}.\) But because \[\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{n^3}{3n^3 + 1} = \frac{1}{3} \ne 0 \cma\] the series does not pass the Alternating Series Test. We therefore cannot use the test to prove convergence, but we also cannot use it to establish divergence. Instead, we see that the limit \[\lim_{n \to \infty} (-1)^n \frac{n^3}{3n^3 + 1}\] does not exist. [As \(n\) grows big, the alternator \((-1)^n\) causes the function to oscillate between \(-1/3\) and \(1/3.\)] So by the Divergence Test, \[\sum_{n = 1}^\infty (-1)^n \frac{n^3}{3n^3 + 1} \divBoxed\]

Alternating Series Error Bound

Consider the infinite, convergent alternating series \[S = \sum_{n = 1}^\infty (-1)^n \, b_n \cmaa b_n \gt 0 \pd\] In most cases, it's impossible to evaluate \(S\) explicitly. Instead, we can estimate \(S\) by using a partial sum \(S_N = \sum_{i = 1}^{N} (-1)^i \, b_i.\) Yet how do we know how accurate our partial sum is? An error bound is the maximum amount by which \(S_N\) is inaccurate in estimating the true sum \(S.\) The Alternating Series Error Bound enables us to bound an error using the absolute value of the first omitted term, \(b_{N + 1}.\) In other words, the value of \(S_N\) differs from the value of \(S\) by no more than \(b_{N + 1}.\) Mathematically, we express this fact as \[\underbrace{\left|S - S_N\right|}_{\textrm{magnitude of error}} \leq \underbrace{b_{N + 1}}_{\abs{\textrm{next term}}}. \] But this fact only applies to alternating series; it is not true for series in general.

ALTERNATING SERIES ERROR BOUND

Consider the convergent alternating series \[S = \sum_{n = 1}^\infty (-1)^n \, b_n \cmaa b_n \gt 0 \pd\] If \(S_N = \sum_{i = 1}^N (-1)^i \, b_i\) is used to estimate \(S,\) then the Alternating Series Error Bound gives \[\abs{S - S_N} \leq b_{N + 1} \pd\]

PROOF The error \(R_N\) is given by \(S - S_N.\) We subtract terms of the partial sum \(S_N\) from the terms of the infinite series \(S \col\) \[ \ba R_N &= S - S_N \nl &= \sum_{n = 1}^\infty (-1)^n \, b_n - \sum_{n = 1}^N (-1)^n \, b_n \nl &= (-1)^{N + 1} \, b_{N + 1} + (-1)^{N + 2} \, b_{N + 2} + (-1)^{N + 3} \, b_{N + 3} + (-1)^{N + 4} \, b_{N + 4} + (-1)^{N + 5} \, b_{N + 5} + \cdots \nl &= (-1)^{N + 1} \par{b_{N + 1} - b_{N + 2} + b_{N + 3} - b_{N + 4} + b_{N + 5} - \cdots} \pd \ea \] Hence, we see \[ \ba \abs{R_N} &= \abs{(-1)^{N + 1} \par{b_{N + 1} - b_{N + 2} + b_{N + 3} - b_{N + 4} + b_{N + 5} - \cdots}} \nl &= b_{N + 1} - b_{N + 2} + b_{N + 3} - b_{N + 4} + b_{N + 5} - \cdots \nl &= b_{N + 1} - \par{b_{N + 2} - b_{N + 3}} - \par{b_{N + 4} - b_{N + 5}} - \cdots \cma \ea \] where each difference in parentheses is positive (or \(0\)) because \(b_{N + 2} \geq b_{N + 3}\) \(\geq b_{N + 4} \geq \cdots.\) Thus, \(\abs{R_N} \leq b_{N + 1}.\) So we conclude that the error in \(S_N\) has a magnitude no greater than \(b_{N + 1}.\) \[\qedproof\]

EXAMPLE 5

The infinite series \(\ds S = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n^2 + 2}\) is approximated by the partial sum \(\ds S_5 = \sum_{i = 1}^5 \frac{(-1)^{i + 1}}{i^2 + 2}.\) Use the Alternating Series Error Bound to find a number \(K\) such that \( \left| S - S_5 \right| \leq K.\)

The sum \(S_5\) uses five terms to estimate \(S.\) The sixth term is the first omitted term, whose magnitude is \[b_6 = \frac{1}{6^2 + 2} = \frac{1}{38} \pd\] Using the Alternating Series Error Bound, we see \[ \abs{S - S_5} \leq b_6 = \underbrace{\boxed{\frac{1}{38}}}_K \]

EXAMPLE 6

The value of the Alternating Harmonic series \[S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{(-1)^{n + 1}}{n} + \cdots\] is approximated by the partial sum \(S_3 = 1 - \frac{1}{2} + \frac{1}{3}.\) Construct an interval in which the value of \(S\) must reside.

Applying the Alternating Series Error Bound, we assert that the value of \(S_3\) differs from \(S\) by no more than the magnitude of the first term we exclude, \(b_4 = 1/4.\) Accordingly, a lower bound to \(S\) is \(\par{S_3 - 1/4}\) and an upper bound to \(S\) is \(\par{S_3 + 1/4}.\) Therefore, \[S_3 - \frac{1}{4} \leq S \leq S_3 + \frac{1}{4} \pd\] Since \(S_3 = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6},\) we have \[S_3 - \frac{1}{4} = \frac{7}{12} \and S_3 + \frac{1}{4} = \frac{13}{12} \pd\] Thus, \[\boxed{\frac{7}{12} \leq S \leq \frac{13}{12}}\]

EXAMPLE 7

The alternating series \(\ds S = \sum_{n = 1}^\infty \frac{(-1)^n}{2\sqrt{n + 3}} \) converges and is approximated by the partial sum \(\ds S_N = \sum_{i = 1}^N \frac{(-1)^i}{2\sqrt{i + 3}} . \) Find the values of \(N\) such that the error of \(S_N\) is no greater than \(0.1.\)

By the Alternating Series Error Bound, an error bound to \(S_N\) is \[\abs{\frac{(-1)^{N + 1}}{2 \sqrt{(N + 1) + 3}}} = \frac{1}{2 \sqrt{N + 4}} \pd\] We equate this expression to be less than or equal to \(0.1,\) finding \[ \frac{1}{2 \sqrt{N + 4}} \leq 0.1 \implies N \geq 21 \pd \] In words, for \(\boxed{N \geq 21}\) the partial sum \(S_N\) differs from \(S\) by no more than \(0.1.\)

Alternating Series An alternating series is a series whose terms alternate in sign. In general, an alternating series takes the form \[\sum_{n = 1}^\infty (-1)^{n + 1} \,b_n \or \sum_{n = 1}^\infty (-1)^n \, b_n \cma\] where \(b_n \gt 0.\) The factor \((-1)^n\) or \((-1)^{n + 1}\) is called an alternator since its sign changes based on whether \(n\) is odd or even. The Alternating Series Test states that \[\sum_{n = 1}^\infty (-1)^n \, b_n \cmaa b_n \gt 0 \cma\] converges if \(\lim_{n \to \infty} b_n = 0\) and \(b_{n + 1} \leq b_n\) for all \(n \geq N\) for some nonnegative integer \(N.\) In words, an alternating series converges if its terms are decreasing in magnitude to \(0.\) But this test cannot prove divergence; usually, we use the Divergence Test to assert that an alternating series is divergent. The Alternating Harmonic series is the sum of alternating reciprocals: \begin{equation} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \cma \eqlabel{eq:alt-harmonic} \end{equation} which converges by the Alternating Series Test.

Alternating Series Error Bound We use the Alternating Series Error Bound to calculate the maximum amount (maximum error) by which a partial sum \(S_N\) differs from the infinite sum \(S.\) Consider the convergent alternating series \[S = \sum_{n = 1}^\infty (-1)^n \, b_n \cmaa b_n \gt 0 \pd\] If \(S_N = \sum_{i = 1}^N (-1)^i \, b_i\) is used to estimate \(S,\) then the Alternating Series Error Bound gives \[\abs{S - S_N} \leq b_{N + 1} \pd\] In words, the magnitude of the error is no greater than the first omitted term in the approximation.