Why the Sign Pattern Matters
A series can have positive and negative terms without being an alternating series in the sense needed for the alternating-series test. A sequence of terms might switch signs repeatedly but have long runs of positive terms and long runs of negative terms. Even if the signs alternate at every step, the series may still fail to converge when the magnitudes do not decrease in the required way.
This distinction matters both when deciding whether an infinite series converges and when estimating the error in a partial sum. The alternating-series test and its remainder bound depend on a specific combination: the signs must alternate, the term magnitudes must be nonincreasing, and those magnitudes must approach zero. Sign changes alone do not supply those conditions.
Strict Alternation and Mere Sign Changes
For a strictly alternating series, the signs can be represented by a factor \((-1)^{n-1}\) or \((-1)^n\). The other factor describes the magnitude. In the usual alternating-series test, the series has the form
where \(b_n\geq 0\). The test applies if \(b_n\) is nonincreasing and \(\lim_{n\to\infty}b_n=0\). A series that begins with a negative term can use \((-1)^n\) instead; changing the first sign does not change the conditions on the magnitudes.
Zero terms do not have a positive or negative sign. If zeros occur, it is safest to consider the sequence of nonzero terms when describing its sign pattern. For applying the standard test and remainder bound, the terms from the relevant point onward must still fit an alternating pattern with nonincreasing magnitudes; a zero term can be included as a zero magnitude if the signed representation remains valid.
The theorem is a sufficient condition, not a claim that every convergent series must alternate. Its conditions describe when this particular test and error estimate are justified.
Why the Alternating-Series Bound Works
The bound comes from cancellation between terms of opposite sign. To see the mechanism, consider a tail whose first term is positive and whose magnitudes \(c_1,c_2,\ldots\) are nonincreasing and tend to zero. Write \(T_m\) for the sum of its first \(m\) terms. Pairing adjacent terms gives the even partial sums
Every parenthesized difference is nonnegative, so \(T_{2m}\geq 0\). These even partial sums increase as more nonnegative pairs are added. The odd partial sums decrease, because passing from \(T_{2m-1}\) to \(T_{2m+1}\) adds the pair \(-c_{2m}+c_{2m+1}\leq 0\). Each odd partial sum is at most \(c_1\): it begins at \(c_1\) and subsequent pairs of the form \(-c_{2j}+c_{2j+1}\) are nonpositive. The even partial sums are also at most \(c_1\), since \(T_{2m}\leq T_{2m+1}\leq c_1\).
The difference between consecutive odd and even partial sums is \(c_{2m+1}\), which tends to zero. The two sequences of partial sums therefore approach the same limit. Since all partial sums lie between zero and \(c_1\), the tail converges to a value between zero and its first magnitude. This proves both the sign statement and the bound for a tail starting with a positive term. If the tail starts negative, changing every sign reverses the sign of the remainder but leaves its absolute value unchanged. Taking the first omitted term as the first term of the tail gives the stated remainder bound.
With alternating signs and nonincreasing magnitudes, each pair cancels partially without changing the sign of the first term in that pair.
The tail's partial sums stay between zero and the magnitude of its first term, so the limiting remainder does too.
If the signs do not alternate or the magnitudes do not decrease, this pairing argument no longer guarantees that the partial tails stay within those bounds. The remainder estimate cannot be inferred just from observing sign changes.
Worked Examples
Worked Example: Infinitely Many Sign Changes Without Convergence
For each positive integer \(k\), form a block consisting of \(k\) copies of \(1/k\), followed by \(k\) copies of \(-1/k\). The blocks begin as
Within each block, the sum of the positive terms is \(k(1/k)=1\), and the sum of the negative terms is \(k(-1/k)=-1\). Thus each complete block has sum \(1-1=0\). At the end of every complete block, the series' partial sum is \(0\).
Halfway through each block, after its \(k\) positive terms, the partial sum is \(0+1=1\), because all preceding blocks have sum zero. There are therefore partial sums equal to \(0\) and partial sums equal to \(1\) arbitrarily far along the sequence. The partial sums do not approach a single limit, so the series diverges.
The terms nevertheless tend to zero: in block \(k\), each term has magnitude \(1/k\), and \(1/k\to 0\). The signs change infinitely often, at the boundaries between positive and negative portions, but they do not strictly alternate; for \(k\geq 2\), the block includes consecutive terms with the same sign. This example shows that repeated sign changes, even together with terms tending to zero, do not ensure convergence.
Check: A complete block contributes \(1-1=0\), while its positive half raises the partial sum to \(1\). These two persistent partial-sum values verify the divergence.
Worked Example: Strict Alternation Without Decreasing Magnitudes
Define the terms in pairs by
The signs strictly alternate, and the magnitudes tend to zero. But they are not nonincreasing. For example, the magnitude of the second term is \(1/2\), while the magnitude of the third is \(1/2\), and more decisively, the magnitude of the fourth is \(1/4\), followed by the larger magnitude \(1/3\) at the fifth term. In general, for \(k>1\), \(1/(k+1)>1/(2k)\), since \(2k>k+1\).
Compute the partial sum through \(2m\) terms by grouping each positive term with the negative term that follows it:
The harmonic partial sums \(\sum_{k=1}^{m}1/k\) grow without bound. For instance, in the group of indices \(2^{j-1}+1\) through \(2^j\), there are \(2^{j-1}\) terms, each at least \(1/2^j\), so that group contributes at least \(1/2\). Adding arbitrarily many such groups makes the partial sums unbounded. Hence \(S_{2m}\) is unbounded and the series diverges.
Check: The pair sum is exactly \(1/(2k)\), so \(S_{2m}=\tfrac12\sum_{k=1}^{m}1/k\). Its divergence confirms that strict sign alternation and terms tending to zero are not enough when magnitudes fail to decrease.
Worked Example: A Remainder Bound Fails for an Irregular Sign Pattern
Consider the series with first three terms \(1,-1/2,-1/3\), and for \(n\geq 4\) define
The remaining tail begins \(1/100,-1/200,1/400,\ldots\). It is a geometric series with first term \(1/100\) and ratio \(-1/2\), so its sum is
The full series converges because it has only three initial terms before this convergent geometric tail. Its sum is
After the first term, the actual remainder is therefore
The next term has magnitude \(1/2\), but \(62/75>1/2\). Thus the alternating-series remainder bound \(|R_1|\leq |a_2|\) would be false here. The sign pattern starts \(+,-,-,+\), so it is not strictly alternating, even though it changes sign infinitely often after the third term.
Check: The computed remainder satisfies \(S_1+R_1=1-62/75=13/75=S\). Since \(62/75>1/2\), this convergent series directly disproves the remainder estimate when the alternating sign condition is absent.
Worked Example: Applying the Bound When Its Conditions Hold
Consider the alternating harmonic series \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\). Here \(b_n=1/n\) is positive, nonincreasing, and tends to zero, so the alternating-series test and remainder bound apply. The fourth partial sum is
The first omitted term is positive and has magnitude \(1/5\). Consequently, the remainder is positive and no greater than \(1/5\):
Check: The bound places the sum between \(7/12\) and \(47/60\), and the error from using \(S_4\) is at most the first omitted magnitude, \(1/5\). The positive direction agrees with the positive fifth term.
A Common Mistake and Its Correction
A common mistake is to treat “the signs change infinitely often” as equivalent to “the terms strictly alternate.” For example, after seeing the series \(1,-1/2,-1/3,1/100,\ldots\), one might claim that its signs alternate and write \(|R_1|\leq 1/2\). But the second and third terms are both negative, so the sign condition fails. The calculation in the previous example gives \(|R_1|=62/75>1/2\), contradicting that claimed bound.
The correction is to inspect the full sequence of signs and magnitudes before applying a test or error estimate. Check that successive nonzero terms have opposite signs, that their magnitudes are nonincreasing, and that those magnitudes tend to zero. If the sequence does not meet these conditions, the alternating-series test and its remainder bound are not justified; use a convergence argument appropriate to the series instead.
Why This Distinction Matters
The alternating-series remainder bound is useful because it turns a partial sum into a controlled approximation: the first omitted magnitude limits the error. But that conclusion is only as reliable as its hypotheses. A series may change signs repeatedly, converge for some other reason, or even have terms tending to zero, without providing the cancellation pattern required by the alternating-series argument.
Check Your Understanding
For each question, distinguish what follows from the sign pattern alone from what requires conditions on the magnitudes.
- What does it mean for the successive nonzero terms of a series to strictly alternate in sign?
- Why does the block series with \(k\) copies of \(1/k\) followed by \(k\) copies of \(-1/k\) diverge, despite its terms tending to zero?
- For the terms \(a_{2k-1}=1/k\) and \(a_{2k}=-1/(2k)\), which alternating-series condition fails, and what are the even partial sums?
- What three conditions justify the alternating-series test and its remainder bound?
- In the series with first terms \(1,-1/2,-1/3\) and a geometric tail, why is the claimed bound \(|R_1|\leq 1/2\) invalid?
