Why Alternating Signs Can Help
In an alternating series, consecutive terms have opposite signs. If their positive magnitudes get smaller, successive terms partially cancel one another. If those magnitudes also approach zero, the later adjustments become arbitrarily small. Together, these facts ensure that the partial sums settle toward a limit.
The alternating-series test turns this idea into a precise criterion. It is a sufficient condition: when its hypotheses hold, the series converges. The conditions are not a claim that every convergent series must have decreasing magnitudes.
The Alternating-Series Test
Let \(a_n\) be a sequence of nonnegative real numbers, and consider the series whose terms alternate in sign, starting with a positive term:
The inequality \(a_{n+1}\leq a_n\) says that the magnitudes are nonincreasing. Equal consecutive values are permitted; strict decrease is not required. The condition \(a_n\geq 0\) identifies \(a_n\) as the magnitude of the \(n\)th term. The limit condition says that these magnitudes eventually become as small as desired.
Check all of the hypotheses before applying the test. Alternating signs by themselves are not enough, and decreasing magnitudes by themselves are not enough if their limit is not zero. When the hypotheses hold, the test establishes convergence, but it does not by itself calculate the value of the infinite sum.
Why the Conditions Guarantee Convergence
We justify the test by showing that sufficiently late partial sums differ by very little. The key observation is that any finite block of consecutive alternating terms has magnitude no greater than the magnitude of its first term. The decreasing condition makes adjacent cancellations predictable.
To see the observation, consider a block starting with a positive sign, with magnitudes \(a_m,a_{m+1},\ldots,a_r\). Pair consecutive terms from the start. Each pair has the form \(a_j-a_{j+1}\), which is nonnegative because \(a_j\geq a_{j+1}\). If the block has an unpaired final term, that term is also nonnegative. Thus the block sum is nonnegative. Pairing from the other end, or separating off the first term and pairing what remains, shows the block sum is at most \(a_m\). A block that starts with a negative sign has the opposite sign pattern, so its absolute value obeys the same bound. Therefore, for every \(r\geq m\),
Now define the \(N\)th partial sum by \(S_N=\sum_{k=1}^{N}(-1)^{k-1}a_k\). If \(q>p\), subtracting two partial sums leaves a consecutive block:
Apply the block bound with \(m=p+1\) and \(r=q\). It gives \(|S_q-S_p|\leq a_{p+1}\). Since \(a_n\to 0\), for every \(\varepsilon>0\) there is an index \(P\) such that \(a_{p+1}<\varepsilon\) whenever \(p\geq P\). Consequently, whenever \(q>p\geq P\),
This is the Cauchy criterion for the sequence of partial sums: for every positive tolerance, all sufficiently late partial sums differ by less than that tolerance. The real numbers are complete, so every Cauchy sequence of real numbers converges. Hence the partial sums converge, which is exactly what it means for the series to converge.
Applying the Test
Worked Example: A Shifted Reciprocal Sequence
Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n+2}\). Here \(a_n=\frac{1}{n+2}\). First, \(a_n>0\) for every \(n\geq1\), since \(n+2>0\).
Next, compare consecutive magnitudes:
because \(n+3>n+2>0\). Thus the sequence is decreasing. Its limit is
since the denominator grows without bound. All the hypotheses are satisfied, so the alternating series converges. As a check, its first terms are \(1/3-1/4+1/5-\cdots\), with positive magnitudes decreasing from one term to the next and tending to zero, just as the test requires.
Worked Example: Geometrically Decreasing Magnitudes
Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{2}{3}\right)^{n-1}\). Set \(a_n=(2/3)^{n-1}\). Since \(2/3>0\), \(a_n>0\). The ratio of consecutive terms is
Therefore \(a_{n+1}<a_n\) for every \(n\). Also, because \(0<2/3<1\), its powers approach zero:
The terms begin \(1-\frac23+\frac49-\frac8{27}+\cdots\), and their magnitudes satisfy all three conditions. The alternating-series test therefore proves convergence. The check is direct: the first magnitude is positive, each new magnitude is \(2/3\) times the previous one, and the magnitudes tend to zero.
Worked Example: Magnitudes Involving a Square Root
Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n+1}}\), so \(a_n=1/\sqrt{n+1}\). Each \(a_n\) is positive because \(n+1>0\). Since \(n+2>n+1\), taking positive square roots gives \(\sqrt{n+2}>\sqrt{n+1}\). Taking reciprocals reverses the inequality, so
Finally, as \(n\) grows, \(\sqrt{n+1}\) grows without bound, and hence \(\lim_{n\to\infty}a_n=0\). The test applies and proves convergence. Checking the conditions again: every magnitude is positive, the displayed inequality verifies decrease, and the reciprocal of an unbounded positive denominator tends to zero.
A Common Pitfall: Signs Alone Do Not Ensure Convergence
It is tempting to think that alternating signs force terms to cancel completely. For example, consider the series with \(a_n=1\):
A mistaken argument might group each \(1-1\) as zero and conclude that the sum is zero. But grouping terms does not make the partial sums converge. The partial sums alternate: \(S_1=1\), \(S_2=0\), \(S_3=1\), and \(S_4=0\). They do not approach a single limit. In this example \(a_n=1\) is nonincreasing, but \(\lim_{n\to\infty}a_n=1\), not zero, so the test cannot be applied.
The correction is to check the hypotheses rather than infer convergence from the signs. A convergent series must have terms tending to zero, and the alternating-series test also requires the magnitudes to be nonincreasing. If either condition fails, this test gives no conclusion.
Using the Test Carefully
A useful way to organize an application is to identify the magnitude \(a_n\) first, then verify each condition explicitly. Keep the sign pattern \((-1)^{n-1}\) separate from the positive magnitude: the test concerns how the magnitudes behave, not merely how the signs alternate.
Write the series as \(\sum(-1)^{n-1}a_n\) and check that \(a_n\geq0\).
Show \(a_{n+1}\leq a_n\) for every relevant index, using an inequality or another valid argument.
Establish that \(a_n\to0\); a sequence can decrease and still approach a positive number.
Once all conditions hold, conclude that the alternating series converges. Do not claim a numerical sum unless it has been separately found.
The key idea is not that positive and negative terms simply disappear in pairs. Instead, decreasing magnitudes make the cancellations orderly, while the zero limit makes the uncancelled part of any sufficiently late block small. The Cauchy argument shows that this control is enough for the partial sums to converge.
Check Your Understanding
For each question, identify which condition of the alternating-series test is relevant.
- State all the hypotheses of the alternating-series test for \(\sum_{n=1}^{\infty}(-1)^{n-1}a_n\).
- Why does the condition \(a_{n+1}\leq a_n\) allow adjacent magnitudes to be paired with nonnegative differences?
- In the convergence proof, why does the bound \(|S_q-S_p|\leq a_{p+1}\) imply that the partial sums form a Cauchy sequence?
- Does the test apply to \(1-\frac12+\frac13-\frac14+\cdots\)? Verify the required conditions.
- Why does the alternating series \(1-1+1-1+\cdots\) fail the test, and what do its partial sums show?
