5.5 Alternating Series

Big idea

Every test so far has required nonnegative terms, because every argument so far has leaned on partial sums that only increase. A series whose signs alternate breaks that setup, and it also behaves better than the positive case rather than worse. Cancellation helps. The alternating harmonic series converges even though the harmonic series does not, and the reason is entirely the alternation.

Picture the partial sums on a number line. Add the first term and you overshoot; subtract the second and you undershoot; add the third and you overshoot by less. If the sizes of the terms decrease steadily, the partial sums bounce with shrinking amplitude, each one trapped between the previous two. If the sizes also go to zero, the bouncing interval closes down on a single point, and that point is the sum. The alternating series test is exactly that picture stated as hypotheses.

The same picture pays a second dividend that no other test in this chapter offers so cheaply. Because the true sum is always caught between consecutive partial sums, the distance from any partial sum to the answer is at most the size of the next term you would have added. That is a usable error bound obtained by looking at a single term, and it is what makes alternating series the workhorses of numerical approximation.

The section also introduces a distinction that will matter for the rest of the course. A series can converge because its terms are small, or it can converge because its signs cancel. Replace every term by its absolute value: if the resulting positive series still converges, the convergence was robust and is called absolute. If the positive series diverges, the convergence depended on the cancellation and is called conditional. The two behave very differently under rearrangement and under the tests that come next.

Decoder

If the sizes of the terms of an alternating series decrease and tend to zero, the series converges, and the error made by stopping after any partial sum is no larger than the size of the first term left out.

Two hypotheses, both needed, both about the sizes $b_n$ rather than the signed terms. The terms must decrease, at least from some index on, and they must go to zero. Neither alone suffices: terms can shrink to zero without decreasing monotonically, and terms can decrease without approaching zero.

The conclusion is one-directional. The test can prove convergence and can never prove divergence. If the hypotheses fail you have learned nothing from this test, and you must reach for something else. In practice, when $b_n$ does not approach zero, the divergence test settles the question immediately, since the signed terms do not approach zero either.

The error statement deserves its own emphasis: the bound is the size of the first omitted term, not of the last included one. A further refinement is free from the same picture: the remainder carries the sign of the first omitted term, so a partial sum ending on a subtraction is an underestimate and one ending on an addition is an overestimate. The classic mistake is applying this bound to a series that is not alternating, or to one whose sizes are not decreasing. Outside those hypotheses it is simply false.

Definitions and results

Alternating series. A series whose terms alternate in sign, written $\sum_{n=1}^{\infty}(-1)^{n-1}b_n$ or $\sum_{n=1}^{\infty}(-1)^{n}b_n$ with $b_n > 0$. The factor $(-1)^{n}$ carries the sign; $b_n$ carries the size.

The alternating series test. If $b_{n+1} \le b_n$ for all large $n$ and $\lim_{n \to \infty} b_n = 0$, then $\sum (-1)^{n-1}b_n$ converges.

Why it converges. The even partial sums increase, the odd partial sums decrease, every even one stays below every odd one, and the gap between $S_{N}$ and $S_{N+1}$ is $b_{N+1} \to 0$. Two monotone bounded sequences closing on each other share a limit.

Error bound. Under the same hypotheses, with $S$ the sum and $S_N$ the $N$th partial sum,

$$ \left|S - S_N\right| \le b_{N+1} $$

and $S - S_N$ has the same sign as the first omitted term.

Absolute convergence. $\sum a_n$ converges absolutely when $\sum |a_n|$ converges. Absolute convergence implies convergence, so proving the positive series converges is enough.

Conditional convergence. $\sum a_n$ converges conditionally when $\sum a_n$ converges but $\sum |a_n|$ diverges. The convergence is then due to cancellation alone.

How to classify. Test $\sum |a_n|$ first, using comparison, the integral test, or a p-series. If it converges you are finished and the answer is absolute. If it diverges, go back to the original series and try the alternating series test; convergence there makes the series conditionally convergent.

Why the distinction matters. Rearranging the order of the terms of an absolutely convergent series never changes its sum. A conditionally convergent series can be rearranged to sum to any number you like, so its value depends on the order in which you add it. That is also why later tests are stated in terms of absolute convergence.

Beyond alternating series. Absolute convergence handles signs that follow no pattern. If $|a_n| \le c_n$ and $\sum c_n$ converges, then $\sum a_n$ converges no matter how its signs behave.

Worked examples

The alternating harmonic series

Decide $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ and classify the convergence.

Here $b_n = \frac1n$. The sizes decrease, since $\frac{1}{n+1} < \frac1n$ for every $n$, and they tend to $0$. Both hypotheses hold, so the series converges.

Now take absolute values: $\sum \frac1n$ is the harmonic series, a p-series with $p = 1$, which diverges. So the convergence is conditional. The whole of it comes from the cancellation between consecutive terms, and none of it from the terms being small.

Stopping early with a guaranteed error

For $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3}$, estimate the sum with four terms and bound the error. Then say how many terms guarantee an error below $0.001$.

The sizes $\frac{1}{n^3}$ decrease to $0$, so the test applies and the error bound is available. Four terms give

$$ S_4 = 1 - \frac18 + \frac{1}{27} - \frac{1}{64} \approx 1 - 0.125 + 0.037037 - 0.015625 = 0.896412 $$

The first omitted term is $+\frac{1}{125} = 0.008$, so the true sum is within $0.008$ of this, and because that omitted term is positive, $S_4$ is an underestimate. The true value is about $0.901543$, which is indeed larger and within $0.0052$ of the estimate, consistent with the bound.

For an error below $0.001$, require $b_{N+1} = \frac{1}{(N+1)^3} < 0.001$, that is $(N+1)^3 > 1000$, that is $N + 1 > 10$. So $N = 10$ terms suffice. Getting three decimal places out of ten terms is respectable, and it is the alternation that makes it possible.

When the test does not apply

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

The sizes are $b_n = \frac{n}{2n+1}$, which increase toward $\frac12$ rather than decreasing to $0$. Both hypotheses fail, so the alternating series test is inconclusive. It is important to stop there and not conclude divergence from the failure of a test.

Something else does settle it. The signed terms $(-1)^{n-1}\frac{n}{2n+1}$ oscillate near $\frac12$ and $-\frac12$, so they do not approach $0$, and the divergence test applies: the series diverges. The lesson is procedural. A failed alternating series test sends you back to the divergence test, not to a conclusion.

Absolute convergence doing the work

Classify $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt n}$ and $\displaystyle\sum_{n=1}^{\infty}\frac{\cos n}{n^2}$.

For the first, $b_n = n^{-1/2}$ decreases to $0$, so the series converges. The absolute series is a p-series with $p = \frac12 \le 1$, which diverges, so the convergence is conditional. Compare this with $\sum \frac{(-1)^{n-1}}{n^2}$, where the absolute series is a p-series with $p = 2$ and the convergence is absolute. Alternation alone does not decide which kind you have; the size of the terms does.

The second series is not alternating at all, since $\cos n$ changes sign irregularly as $n$ runs through the integers. The alternating series test is unavailable. But $|\cos n| \le 1$ gives

$$ \left|\frac{\cos n}{n^2}\right| \le \frac{1}{n^2} $$

and $\sum \frac{1}{n^2}$ converges by the p-series result, so $\sum \left|\frac{\cos n}{n^2}\right|$ converges by direct comparison. The series therefore converges absolutely, hence converges. Absolute convergence is the tool that handles signs with no pattern to exploit.

Practice

One skill covering the section. Expect to verify both hypotheses before applying the test, to produce an error bound or a required number of terms from the first omitted term, and to classify convergence as absolute, conditional, or absent.

Quiz

Five items on the alternating series test and its hypotheses, the remainder bound, and telling absolute from conditional convergence.