Classifying Alternating-Series Examples
A series can alternate in sign and still behave in several different ways. It may diverge because its terms do not approach zero; it may converge only because of cancellation; or it may converge absolutely, whether or not the standard alternating-series remainder estimate applies.
These are separate questions. First determine whether the series converges. If it does, test whether the series of absolute values converges. Then check the term magnitudes separately to decide whether the alternating-series remainder estimate is available.
A series \(\sum a_n\) converges absolutely if \(\sum |a_n|\) converges. It converges conditionally if \(\sum a_n\) converges but \(\sum |a_n|\) diverges. These descriptions apply to convergent series; a divergent series is neither absolutely nor conditionally convergent.
Three Checks, Kept Separate
For a series written as \(\sum (-1)^{n-1}b_n\), the convergence test and the error estimate use related but distinct conditions. The Alternating-Series Test says that if \(b_n\) is nonincreasing from some index onward and \(\lim_{n\to\infty}b_n=0\), then the alternating series converges. A finite number of initial terms do not affect convergence.
The standard alternating-series remainder estimate uses the same eventual monotonicity and zero-limit conditions. If the conditions hold from index \(m\) onward, then for \(N\ge m\), the remainder after the \(N\)th partial sum satisfies \(|R_N|\le b_{N+1}\). The estimate is about the error in approximating the sum by that partial sum. It does not require absolute convergence.
Why the Remainder Estimate Works
Here is a justification of the estimate. Suppose \(b_n\) is nonincreasing and tends to zero from index \(m\) onward. The Alternating-Series Test gives convergence, so the remainder after \(N\ge m\) is the limit of the finite alternating tails. After factoring out the sign of the first omitted term, these tails have the form
For even \(r\), grouping adjacent terms shows that \(T_r\ge0\), because each difference such as \(b_{N+1}-b_{N+2}\) is nonnegative. Also,
For odd \(r\), grouping from the second term gives
so \(T_r\le b_{N+1}\). Grouping from the first term shows \(T_r\ge0\). Thus every finite tail lies between zero and the first omitted magnitude. Taking the limit preserves those inequalities. Restoring the possible sign of the first omitted term gives
This argument also explains why the bound is tied to a decreasing sequence of magnitudes: the paired differences must be nonnegative.
Worked Classifications
Worked Example: Conditional Convergence with Decreasing Magnitudes
Classify \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+1}\), and decide whether its alternating-series remainder estimate applies.
The magnitudes are \(b_n=\frac{n}{n^2+1}\). To check decrease, compare consecutive terms:
For every integer \(n\ge1\), the numerator \(n^2+n-1\) is positive and the denominator is positive. Hence \(b_n>b_{n+1}\). Also, \(0<b_n\le 1/n\), so \(b_n\to0\). The Alternating-Series Test therefore gives convergence.
For absolute convergence, compare the magnitudes with the harmonic series. For \(n\ge1\),
The comparison series \(\sum \frac{1}{2n}\) diverges, so the series of magnitudes diverges. The original series is therefore conditionally convergent. Since its magnitudes decrease to zero, the remainder estimate applies for every \(N\ge1\):
As a check on the classification: the magnitude differences are positive, the magnitudes tend to zero, and the absolute-value comparison proves divergence of the magnitude series. Those facts establish conditional convergence and the stated estimate conditions.
Worked Example: Absolute Convergence and a Valid Remainder Bound
Classify \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n^2}\), and find a remainder bound after the second term.
Here \(b_n=1/n^2\). It decreases because \(1/n^2>1/(n+1)^2\) for every \(n\ge1\), and \(b_n\to0\). Thus the alternating-series conditions hold. The series also converges absolutely: \(\sum |(-1)^{n-1}/n^2|=\sum 1/n^2\), which converges by the \(p\)-series test with \(p=2>1\).
The second partial sum is
The first omitted magnitude is \(b_3=1/3^2=1/9\). Since the magnitudes decrease to zero, the remainder after \(S_2\) satisfies
The check is that \(b_1>b_2>b_3>\cdots>0\), with limit zero, and that the magnitude series converges. This example is absolutely convergent and still satisfies the alternating-series remainder estimate.
Worked Example: Absolute Convergence Without the Alternating-Series Remainder Estimate
Consider \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{2+(-1)^n}{n^2}\). Classify it and determine whether the standard alternating-series remainder estimate applies.
The magnitudes are \(b_n=\frac{2+(-1)^n}{n^2}\). Since \(-1\le(-1)^n\le1\), they satisfy \(0<b_n\le3/n^2\). Therefore
The series converges absolutely. But its magnitudes are not nonincreasing, even after discarding any finite number of terms. For odd \(n\), \(b_n=1/n^2\), while \(b_{n+1}=3/(n+1)^2\). For odd \(n\ge3\),
The final inequality holds for every odd integer \(n\ge3\). Thus the magnitudes increase at infinitely many consecutive pairs. The standard alternating-series remainder estimate does not apply to this series, even though it converges absolutely. The check is precisely the distinction between the two conditions: comparison proves absolute convergence, while infinitely many increases rule out eventual nonincrease.
Worked Example: Magnitudes That Decrease Only Eventually
Classify \(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n}{n^2+4}\), and specify when its remainder estimate can be used.
Set \(b_n=n/(n^2+4)\). The first two magnitudes are
so the magnitudes initially increase. For \(n\ge2\), however,
The denominator is positive, and \(n^2+n-4\ge2\) for \(n\ge2\). Thus \(b_n>b_{n+1}\) from \(n=2\) onward. Also, \(b_n\le1/n\), so \(b_n\to0\). The alternating series converges by the Alternating-Series Test, which permits finitely many initial increases.
For \(n\ge2\), \(n^2+4\le2n^2\), and hence \(b_n\ge1/(2n)\). The harmonic comparison shows that the magnitude series diverges, so the original series is conditionally convergent. The remainder estimate applies after partial sums with \(N\ge2\), not to the first partial sum using the standard monotonicity condition from its start. For example,
The check is that the increase occurs only between the first two magnitudes, while all later magnitudes decrease to zero. Therefore convergence and the estimate are justified for the tail beginning with \(b_3\); absolute convergence fails by harmonic comparison.
A Divergence Check Before Using Alternation
Alternating signs do not guarantee convergence. For example, in \(\sum_{n=1}^{\infty}(-1)^{n-1}(1+1/n)\), the term magnitudes tend to \(1\), not zero. A necessary condition for convergence of any series \(\sum a_n\) is \(a_n\to0\). Here that condition fails, so the series diverges. Neither the Alternating-Series Test nor its remainder estimate applies.
Why the Distinctions Matter
Convergence tells whether a series has a finite sum. Absolute or conditional status describes how convergence relates to the magnitudes. The alternating-series remainder estimate answers a different question: whether a particular partial sum has a certified error bound from the first omitted magnitude. Absolute convergence alone does not supply that alternating-series bound, as the nonmonotone example shows. Conversely, conditional convergence can come with the bound whenever the magnitudes decrease to zero, as in the first and fourth examples.
For each new example, write down three conclusions separately: convergence status, absolute or conditional status when applicable, and remainder-estimate applicability. This prevents a correct observation about one property from being mistaken for a conclusion about another.
Check Your Understanding
For each question, state which conditions you would check before giving a classification.
- What condition on the terms can prove that an alternating-sign series diverges immediately?
- If an alternating series satisfies the Alternating-Series Test, what additional test distinguishes absolute from conditional convergence?
- Can the alternating-series remainder estimate apply to an absolutely convergent series? Explain which conditions decide.
- If the magnitudes increase for the first two terms but decrease thereafter to zero, what does that imply for convergence and for remainder bounds on sufficiently late partial sums?
- Why does absolute convergence by comparison not, by itself, establish the standard alternating-series remainder estimate?
