From Finite Total Magnitude to a Signed Sum
In the previous tutorial, absolute convergence was defined by convergence of the series of term magnitudes. That definition deliberately ignores signs: it asks whether the terms have a finite total size before any cancellation is taken into account. We now establish the central consequence of this stronger condition. If the magnitudes have a finite total, then the original signed series must converge.
The reason is that the absolute values control every finite block of signed terms. By the triangle inequality, the magnitude of a signed block cannot exceed the sum of the magnitudes in that block. Since absolute convergence makes all sufficiently late blocks of magnitudes small, the signed partial sums satisfy the Cauchy condition as well.
Proof. Let \(\varepsilon>0\). Since \(\sum |a_n|\) converges, the Cauchy Criterion for Absolute Convergence gives an integer \(N\) such that, whenever \(q\geq p\geq N\), $$ \sum_{n=p}^{q}|a_n|<\varepsilon. $$ The finite triangle inequality then gives $$ \left|\sum_{n=p}^{q}a_n\right| \leq \sum_{n=p}^{q}|a_n| <\varepsilon. $$ Thus, for every \(\varepsilon>0\), all sufficiently late finite blocks of the signed series have magnitude less than \(\varepsilon\). By the Cauchy Criterion for Series, \(\sum a_n\) converges. \(\square\)
This proof makes clear which condition is doing the work. The signed terms do not need to alternate, and they do not need to follow any regular sign pattern. The only input is that the total magnitude in every sufficiently late block is small. Absolute convergence therefore guarantees convergence without relying on cancellation.
Worked Examples
Worked Example: A Geometric Series with Alternating Signs
Consider $$ \sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{2}{5}\right)^n. $$ The magnitudes form the geometric series $$ \sum_{n=1}^{\infty}\left(\frac{2}{5}\right)^n. $$ Its ratio is \(2/5\), whose absolute value is less than \(1\), so it converges. The series in the example is therefore absolutely convergent, and the theorem shows that it converges.
We can also identify its sum. Its first term is \(2/5\), and its common ratio is \(-2/5\). The geometric-series formula gives $$ \sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{2}{5}\right)^n =\frac{2/5}{1-(-2/5)} =\frac{2/5}{7/5} =\frac{2}{7}. $$ For comparison, the sum of the magnitudes is $$ \sum_{n=1}^{\infty}\left(\frac{2}{5}\right)^n =\frac{2/5}{1-2/5} =\frac{2/5}{3/5} =\frac{2}{3}. $$ Both sums are finite; the signed sum is smaller because its terms alternate.
Worked Example: Irregular Signs and a Quadratic Denominator
Consider the series $$ \sum_{n=1}^{\infty}\frac{(-1)^{\lfloor\sqrt n\rfloor}}{n^2+n}, $$ where \(\lfloor\sqrt n\rfloor\) denotes the greatest integer less than or equal to \(\sqrt n\). The sign changes according to blocks of indices, so it need not alternate term by term. For every \(n\geq1\), the denominator is positive and $$ \left|\frac{(-1)^{\lfloor\sqrt n\rfloor}}{n^2+n}\right| =\frac{1}{n^2+n} \leq\frac{1}{n^2}, $$ because \(n^2+n\geq n^2\). The comparison series \(\sum 1/n^2\) converges by the p-Series Convergence Criterion, with \(p=2>1\). The Comparison Test for Absolute Convergence shows that the displayed series is absolutely convergent. Hence it converges, regardless of the irregular arrangement of its signs.
Worked Example: A Remainder Estimate from Absolute Convergence
Consider $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}. $$ The magnitudes can be written as $$ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}. $$ For every positive integer \(M\), the finite sum of magnitudes therefore telescopes: $$ \sum_{n=1}^{M}\frac{1}{n(n+1)} =\sum_{n=1}^{M}\left(\frac{1}{n}-\frac{1}{n+1}\right) =1-\frac{1}{M+1}. $$ As \(M\) tends to infinity, this expression tends to \(1\). Thus the series of magnitudes converges, and the signed series converges by the theorem.
For this example, the absolute tail after the \(N\)th term is $$ \sum_{n=N+1}^{\infty}\frac{1}{n(n+1)} =\frac{1}{N+1}. $$ Consequently, if \(S\) denotes the sum and \(S_N\) the \(N\)th partial sum, the absolute tail estimate proved below gives $$ |S-S_N|\leq\frac{1}{N+1}. $$ For instance, after \(N=9\) terms, the error is at most \(1/10\). The estimate does not require evaluating the signed sum.
Absolute Tails Control the Error
The convergence theorem has a quantitative consequence. Once the signed series has a sum, the error after finitely many terms is no greater than the total magnitude of the omitted terms. This is stronger than merely knowing that the partial sums approach a limit: it provides a direct bound that can be estimated using positive-series methods.
Proof. By Absolute Convergence Implies Convergence, the signed series converges to \(S\). For each integer \(M>N\), the finite triangle inequality gives $$ |S_M-S_N| =\left|\sum_{n=N+1}^{M}a_n\right| \leq\sum_{n=N+1}^{M}|a_n| \leq\sum_{n=N+1}^{\infty}|a_n|. $$ The final quantity is finite and does not depend on \(M\). As \(M\) tends to infinity, \(S_M-S_N\) tends to \(S-S_N=R_N\). Continuity of the absolute value therefore gives $$ |R_N| =\lim_{M\to\infty}|S_M-S_N| \leq\sum_{n=N+1}^{\infty}|a_n|. $$ This proves the bound. \(\square\)
The right-hand side is a positive tail, so any suitable estimate for \(\sum |a_n|\) can be used to control the error. In the preceding example, telescoping gave that tail exactly. In other cases, comparison with a geometric series or a p-series may provide a convenient upper bound.
Why the Converse Fails
The theorem goes in one direction only: absolute convergence implies convergence, but convergence alone does not imply absolute convergence. A signed series may converge because of cancellation even though the magnitudes have an infinite sum. The following example makes the distinction explicit without using a term-by-term alternating pattern from the previous tutorial.
Worked Example: Convergence Without Absolute Convergence
Consider $$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt n}. $$ The sequence \(b_n=1/\sqrt n\) is positive and decreasing, since \(b_{n+1}\leq b_n\) follows from \(n+1\geq n\). Also \(b_n\) tends to zero. The Alternating Series Test therefore shows that the signed series converges.
Its series of magnitudes, however, is $$ \sum_{n=1}^{\infty}\frac{1}{\sqrt n} =\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}. $$ This is a p-series with \(p=1/2\leq1\), so it diverges by the p-Series Convergence Criterion. The original series thus converges but is not absolutely convergent; it is conditionally convergent. The convergence comes from cancellation, not from a finite total of magnitudes.
This distinction matters when choosing a test. If the magnitude series converges, the theorem immediately settles convergence of the signed series, and the remainder bound is available as well. If the magnitude series diverges, no conclusion about the signed series follows from that fact alone: one must then examine the signs and any cancellation separately.
Determine whether \(\sum |a_n|\) converges, using comparison, a geometric bound, or another positive-series test.
If the magnitude series converges, conclude that \(\sum a_n\) converges by Absolute Convergence Implies Convergence.
Bound \(|S-S_N|\) by the positive tail \(\sum_{n=N+1}^{\infty}|a_n|\).
Check Your Understanding
Use the theorem and examples to answer the following questions.
- Which condition on \(\sum |a_n|\) guarantees that \(\sum a_n\) converges?
- How does the Cauchy Criterion for Absolute Convergence lead to the Cauchy condition for the signed series?
- Why does an irregular pattern of signs cause no difficulty when the series is absolutely convergent?
- What upper bound does absolute convergence give for \(|S-S_N|\)?
- Why does convergence of the alternating series with terms \((-1)^{n-1}/\sqrt n\) not imply absolute convergence?