Why Look at the Magnitudes?
A series can have both positive and negative terms. Those signs may cause cancellation: a positive term can be partly or completely offset by a negative one. This can make the original series converge even when its terms do not all contribute in the same direction.
To study convergence without relying on cancellation, examine the size of each term. For a real series \(\sum_{n=1}^{\infty} a_n\), its series of term magnitudes is \(\sum_{n=1}^{\infty} |a_n|\). This is a series with nonnegative terms, so it records the total size of the terms without regard to their signs.
The definition asks about the convergence of a new series, not merely whether the original series has a finite sum. The key result is that absolute convergence is a strong enough condition to guarantee convergence of the original series.
The Main Result
The proof uses the Cauchy criterion for series: a series converges if and only if, for every \(\varepsilon>0\), there is an index \(N\) such that every finite tail \(\sum_{k=n+1}^{m} a_k\), with \(m>n\ge N\), has absolute value less than \(\varepsilon\). This criterion expresses convergence in terms of the tails of the partial sums.
Because \(\sum |a_n|\) converges, its tails can be made arbitrarily small.
The triangle inequality shows that the absolute value of a tail of \(\sum a_n\) is no larger than the corresponding tail of \(\sum |a_n|\).
The original partial sums satisfy the criterion, so the original series converges.
Here is the complete argument. Let \(\varepsilon>0\). Since \(\sum_{n=1}^{\infty}|a_n|\) converges, the Cauchy criterion gives an index \(N\) such that whenever \(m>n\ge N\),
For the same indices, the triangle inequality gives
Thus every sufficiently late finite tail of the original series has absolute value less than \(\varepsilon\). Its partial sums satisfy the Cauchy criterion. Since the real numbers are complete, a Cauchy sequence of real partial sums converges. Therefore \(\sum a_n\) converges, as claimed.
The argument does not require the terms to alternate, or to follow any particular sign pattern. It uses only convergence of the magnitude series and the triangle inequality.
Worked Examples
Worked Example: An Alternating Geometric Series
Determine whether \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{2^n}\) is absolutely convergent, and find its sum.
First take the magnitude of each term:
The magnitude series is geometric, with first term \(1/2\) and common ratio \(1/2\). Since the ratio has absolute value less than \(1\), its sum is
Therefore the original series is absolutely convergent. To find its sum, its first term is \(1/2\) and its common ratio is \(-1/2\). The geometric-series formula gives
Check: the \(N\)th partial sum is \(\frac{1}{3}\bigl(1-(-1/2)^N\bigr)\). Since \((-1/2)^N\) tends to zero, these partial sums tend to \(1/3\), as calculated.
Worked Example: A Rational-Term Series
Show that \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2+4}\) is absolutely convergent.
The magnitude series is
Compare its terms with \(1/n^2\). The limit of their ratio is
The limit comparison test says that if two positive-term series have a term ratio tending to a positive finite number, then either both series converge or both diverge. The \(p\)-series \(\sum 1/n^2\) converges because \(p=2>1\). The ratio limit is \(1\), which is positive and finite, so \(\sum 1/(n^2+4)\) converges as well. This is precisely the magnitude series; hence the original series is absolutely convergent and, by the theorem, convergent.
Check: every magnitude term is positive, and the comparison test establishes convergence of their sum. The absolute-convergence definition is therefore satisfied.
Worked Example: A Telescoping Magnitude Series
Determine whether \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+1)}\) is absolutely convergent.
Taking magnitudes removes the sign factor:
Use partial fractions, since \(1/n-1/(n+1)=1/[n(n+1)]\). The \(N\)th partial sum of the magnitude series is therefore
As \(N\) tends to infinity, \(1/(N+1)\) tends to zero, so these partial sums tend to \(1\). The magnitude series converges, and the original series is absolutely convergent. The theorem then guarantees that the original series converges.
Check: the first three magnitude partial sums are \(1/2\), \(1/2+1/6=2/3\), and \(2/3+1/12=3/4\), agreeing with \(1-1/(N+1)\) for \(N=1,2,3\). The formula tends to \(1\), as required.
What the Magnitude Series Tells You
Absolute convergence is useful because it gives information about the original series without needing to track the effects of positive and negative terms. The central inequality is
The right-hand side measures the total magnitude in a finite tail. The left-hand side measures the net tail after any cancellation. The net tail cannot be larger than the total magnitude.
A related consequence gives a bound on the error after truncating an absolutely convergent series. If \(S=\sum_{n=1}^{\infty}a_n\) and \(S_N=\sum_{n=1}^{N}a_n\), then
To obtain this inequality, apply the finite-tail triangle inequality through an index \(m>N\), then let \(m\) tend to infinity. The finite sums on the left tend to \(S-S_N\), and the magnitude tails tend to \(\sum_{n=N+1}^{\infty}|a_n|\). Thus the tail of the magnitude series supplies an upper bound for the error in the original sum.
A Common Mistake: Taking the Absolute Value After Summing
Absolute convergence does not mean taking the absolute value of the final sum. The absolute values must be taken term by term before summing. For instance, for the two terms \(1\) and \(-1\), taking the absolute value after adding gives
whereas adding the term magnitudes gives
These are different quantities because cancellation occurs in the first calculation but not the second. For an infinite series, the definition concerns \(\sum |a_n|\), not \(\left|\sum a_n\right|\). A finite total sum, by itself, does not establish absolute convergence.
Summary
To test for absolute convergence, form the series \(\sum |a_n|\) and determine whether that series converges. If it does, then the original series converges: every tail of the original series is bounded in magnitude by the corresponding tail of the magnitude series. The same inequality also gives a useful bound on the truncation error.
Check Your Understanding
Answer each question using the definition and results developed here.
- What series must converge for \(\sum a_n\) to be absolutely convergent?
- Why does convergence of the magnitude series imply that the partial sums of \(\sum a_n\) satisfy the Cauchy criterion?
- Find the first \(N\) partial sums of \(\sum_{n=1}^{N}1/[n(n+1)]\) in a formula, and state their limit.
- For an absolutely convergent series with sum \(S\), what quantity bounds \(|S-S_N|\)?
- Explain why \(\left|\sum a_n\right|\) is not the definition of absolute convergence.
