When the Alternating-Series Test Is Inconclusive

When a Test Gives No Conclusion

The Alternating-Series Test is useful because it gives a straightforward way to prove convergence for many series with alternating signs. Its conditions are sufficient, however, not necessary: a series can converge even when those conditions fail. It is therefore important to distinguish a failed test from a failed series.

For an alternating series, the crucial first question is whether its terms approach zero. If they do not, the series diverges. If they do, but the magnitudes are not eventually decreasing, the Alternating-Series Test gives no conclusion. The series might still converge, or it might diverge; a different argument is needed.

The Test and What Its Failure Means

Alternating-Series Test: Suppose \(b_n\geq 0\), and consider \(\sum_{n=1}^{\infty}(-1)^{n+1}b_n\). If \(b_n\) is eventually nonincreasing and \(\lim_{n\to\infty}b_n=0\), then the alternating series converges. “Eventually nonincreasing” means there is an index \(N\) such that \(b_{n+1}\leq b_n\) for every \(n\geq N\).

The two conditions do different work. The zero-limit condition says the individual terms become small. The decrease condition controls how successive positive and negative terms offset one another. If the decrease condition fails, the test's conclusion is unavailable; that failure alone says nothing about whether the series converges.

By contrast, failure of the zero-limit condition is decisive. For any convergent series, its terms must tend to zero. To see why, let \(S_n\) be its \(n\)th partial sum. The \(n\)th term is the difference between two consecutive partial sums:

a_n=S_n-S_{n-1}.

If the series converges, then \(S_n\) and \(S_{n-1}\) both approach the same limit \(S\). Their difference must therefore approach zero: \(a_n\to S-S=0\). This is the term test for divergence, also called the \(n\)th-term test: if \(\lim_{n\to\infty}a_n\neq 0\), or if that limit does not exist, the series diverges.

Key distinction: If the magnitudes \(b_n\) do not approach zero, the alternating series diverges by the term test. If \(b_n\to 0\) but is not eventually nonincreasing, the Alternating-Series Test is inconclusive; another method must decide the question.

Worked Example: Terms That Do Not Approach Zero

Worked Example: A Nonzero Limit Forces Divergence

Consider \(\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n+1}\). Its magnitudes are \(b_n=\frac{n}{n+1}\). Compute their limit:

\lim_{n\to\infty}b_n = \lim_{n\to\infty}\frac{n}{n+1} = \lim_{n\to\infty}\frac{1}{1+1/n} = 1.

Thus the series terms \(a_n=(-1)^{n+1}b_n\) do not approach zero: their magnitudes approach \(1\). More explicitly, the odd-indexed terms approach \(1\), while the even-indexed terms approach \(-1\). In particular, the terms do not have limit zero.

The term test proves that the series diverges. This conclusion does not require checking whether the magnitudes decrease: they increase toward \(1\), and, more importantly, the necessary zero-limit condition fails.

Check: The conclusion agrees with the necessary condition for convergence. Since \(\lim a_n\neq 0\), the series cannot converge.

When the Magnitudes Tend to Zero but Do Not Decrease

Suppose \(b_n\to 0\), but the magnitudes rise at infinitely many steps. Then the term test does not establish divergence, and the Alternating-Series Test cannot be applied. It is not enough to observe that the terms alternate and approach zero: their pattern of sizes matters, and a separate argument is needed.

Pairing consecutive terms can be helpful. For a series written as \(\sum_{n=1}^{\infty}(-1)^{n+1}b_n\), its first \(2N\) terms can be grouped into \(N\) pairs. Grouping in this fixed order does not change the partial sums at those even indices. If those even partial sums have a finite limit, and the intervening odd partial sums differ by a term that approaches zero, then the full sequence of partial sums has that same limit.

Worked Example: Convergence Despite Repeated Increases

Worked Example: Pairing Reveals Convergence

Define the magnitudes, for \(k\geq 1\), by

b_{2k-1}=\frac{1}{k}, \qquad b_{2k}=\frac{1}{k}+\frac{1}{k^2}.

The magnitudes tend to zero: both \(\frac{1}{k}\) and \(\frac{1}{k}+\frac{1}{k^2}\) approach zero as \(k\to\infty\). But within every pair they increase, since

b_{2k}-b_{2k-1} = \left(\frac{1}{k}+\frac{1}{k^2}\right)-\frac{1}{k} = \frac{1}{k^2}>0.

So the sequence is not eventually nonincreasing, and the Alternating-Series Test does not apply. Now examine the alternating series by pairing its terms:

(-1)^{2k-2}b_{2k-1}+(-1)^{2k-1}b_{2k} = \frac{1}{k}-\left(\frac{1}{k}+\frac{1}{k^2}\right) = -\frac{1}{k^2}.

Consequently, the even partial sums are

S_{2N} = \sum_{k=1}^{N}\left(-\frac{1}{k^2}\right) = -\sum_{k=1}^{N}\frac{1}{k^2}.

The \(p\)-series \(\sum_{k=1}^{\infty}1/k^2\) converges, because its exponent is \(p=2>1\). Thus the even partial sums approach a finite limit. The next odd partial sum differs from the preceding even one by

S_{2N+1}-S_{2N} = b_{2N+1} = \frac{1}{N+1},

and \(\frac{1}{N+1}\to 0\). Hence the odd partial sums approach the same limit as the even partial sums, so the series converges.

Check: Pairing gives exactly \(-1/k^2\) for the \(k\)th pair, and the unpaired term after \(2N\) terms is exactly \(1/(N+1)\). The pair series converges and the leftover term vanishes, verifying convergence even though the magnitudes increase within every pair.

Worked Example: Vanishing Terms but Divergent Series

Worked Example: Pairing Reveals Divergence

For \(k\geq 1\), set

b_{2k-1}=\frac{2}{k}, \qquad b_{2k}=\frac{1}{k}.

Both subsequences tend to zero, so \(b_n\to 0\). The magnitudes are not eventually nonincreasing: for \(k>1\),

b_{2k+1}-b_{2k} = \frac{2}{k+1}-\frac{1}{k} = \frac{2k-(k+1)}{k(k+1)} = \frac{k-1}{k(k+1)} > 0.

Thus there are increases at arbitrarily large indices. The Alternating-Series Test is inconclusive. Pair the signed terms instead:

(-1)^{2k-2}b_{2k-1}+(-1)^{2k-1}b_{2k} = \frac{2}{k}-\frac{1}{k} = \frac{1}{k}.

It follows that the even partial sums are

S_{2N} = \sum_{k=1}^{N}\frac{1}{k}.

These are the partial sums of the harmonic series, which diverges. Therefore the even partial sums of the alternating series are unbounded, so the alternating series diverges.

Check: Every pair contributes \(1/k\), not a term small enough to make the cumulative paired sums finite. The pair calculation agrees with the divergent harmonic partial sums, confirming divergence even though the individual terms tend to zero.

A Reliable Diagnostic Process

When the Alternating-Series Test appears not to apply, use the following order. It prevents the common mistake of treating every failed condition as evidence of divergence.

Write the signed term and its magnitude.
Express the series as \(\sum(-1)^{n+1}b_n\) or \(\sum(-1)^n b_n\), with \(b_n\geq 0\).

Check the limit of the terms.
If the terms do not tend to zero, conclude divergence by the term test. If they do, continue.

Check eventual decrease.
Look for one index after which every successive magnitude is no larger than the previous one. A finite number of early increases does not prevent eventual decrease.

If decrease fails, change methods.
Consider pairing consecutive terms, or use a suitable comparison or other convergence test established earlier in the course. Do not infer a conclusion from the failed condition alone.

Common Mistake: Treating Inconclusive as Divergent

A frequent error is to notice that the magnitudes are not decreasing and immediately write “the series diverges.” That conclusion does not follow. For example, in the convergent example above, \(b_{2k}>b_{2k-1}\) for every \(k\), yet the series converges because the paired terms sum to \(-1/k^2\).

The correction is to ask what exactly failed. If the magnitudes do not tend to zero, the term test proves divergence. If they tend to zero but are not eventually nonincreasing, the Alternating-Series Test is simply silent. The convergent and divergent examples above both have magnitudes tending to zero and failing eventual decrease, but their paired sums behave differently.

Why This Distinction Matters

A convergence test is a tool with stated hypotheses, not a rule that classifies every series. The Alternating-Series Test offers a useful guarantee when its conditions hold. Outside those conditions, there is no automatic conclusion. Pairing may reveal cancellation and convergence, or it may expose a divergent series of paired contributions. Other established tests can also be appropriate, depending on the terms.

This discipline is useful well beyond alternating series: always separate the result a test guarantees from what happens when a hypothesis is absent. A missing sufficient condition does not establish the opposite outcome. Only a valid argument for convergence or divergence can settle the series.

Check Your Understanding

For each question, identify what the stated information establishes—and what still requires another argument.

If the magnitudes of an alternating series approach \(3\), what can you conclude, and which result justifies it?
If the magnitudes tend to zero but increase at infinitely many indices, what does the Alternating-Series Test establish?
In the convergence example, what is the signed sum of the terms numbered \(2k-1\) and \(2k\)?
In the divergence example, what are the even partial sums, and why do they rule out convergence?
Why is a finite number of increases at the start compatible with the eventual-decrease condition?