Why Compare Absolute Values?
A series may contain both positive and negative terms. Those signs can make its partial sums behave differently from the partial sums of the absolute values. To test whether a series converges absolutely, we remove the signs and study a positive-term series instead.
This gives a systematic way to use familiar benchmarks, especially geometric series and p-series. Direct comparison is useful when an inequality is clear; limit comparison is useful when two positive terms have the same long-term scale.
If the absolute-value series diverges, the original series does not converge absolutely. That conclusion alone does not determine whether the original signed series converges; it only answers the question of absolute convergence.
Comparison Tools for Absolute Convergence
To apply comparison, first form the nonnegative terms \(a_n=|u_n|\). Then compare \(a_n\) with a known positive-term series. The comparisons below may be applied from some index onward: changing finitely many terms does not change whether a series converges.
The direction of the inequality matters: a smaller nonnegative term is bounded by a series whose total sum is finite. Common benchmarks are \(\sum 1/n^p\), which converges when \(p>1\), and geometric series with ratio of absolute value less than \(1\).
Then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. For absolute convergence, use \(a_n=|u_n|\) and choose a positive benchmark \(b_n\) with a known series behavior.
The limit condition says that, for large \(n\), \(a_n\) and \(b_n\) differ by factors bounded above and below by positive constants. More precisely, since \(a_n/b_n\to L>0\), eventually
Multiplying by \(b_n>0\) gives \((L/2)b_n<a_n<(3L/2)b_n\). Thus each series is bounded by a constant multiple of the other in its tail, which explains why they share convergence or divergence.
Why Absolute Convergence Guarantees Convergence
Absolute convergence is a strong conclusion: it guarantees that the original series converges, regardless of the arrangement of positive and negative terms in its usual order. The reason is that the absolute values bound every possible tail sum.
Here is a proof using 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 beginning after \(N\) has absolute value less than \(\varepsilon\).
Since \(\sum |u_n|\) converges, its tails approach zero. Given \(\varepsilon>0\), choose \(N\) so that \(\sum_{n=N+1}^{\infty}|u_n|<\varepsilon\).
For integers \(q\geq p>N\), the triangle inequality gives \(\left|\sum_{n=p}^{q}u_n\right|\leq\sum_{n=p}^{q}|u_n|\leq\sum_{n=N+1}^{\infty}|u_n|<\varepsilon\).
Every sufficiently late finite tail of \(\sum u_n\) has absolute value less than \(\varepsilon\). Since this holds for every \(\varepsilon>0\), the series \(\sum u_n\) converges.
This proof also shows why comparison of absolute values is useful: a convergent bound on \(\sum |u_n|\) controls the tails of the original series directly.
Worked Examples
Worked Example: Direct Comparison for an Alternating Rational Series
Determine whether the series
converges absolutely.
Step 1: Take absolute values. For every positive integer \(n\),
because \(4n+1>0\), the denominator is positive, and \(|(-1)^{n+1}|=1\).
Step 2: Bound the numerator above and the denominator below. Since \(n\geq1\), \(1\leq n\), so \(4n+1\leq5n\). Also \(n^3+2n^2+5\geq n^3>0\). Therefore,
The benchmark \(\sum 5/n^2=5\sum 1/n^2\) converges because it is a constant multiple of the p-series with \(p=2>1\). Direct comparison proves that the series of absolute values converges.
Conclusion and check: The given series converges absolutely, so the theorem above guarantees that the original series converges. The comparison has the required direction, \(0\leq |u_n|\leq5/n^2\), and its benchmark is convergent.
Worked Example: Limit Comparison Shows Absolute Divergence
Determine whether the series
converges absolutely.
Step 1: Take absolute values and choose a benchmark. The absolute-value term is
It has the same long-term scale as \(1/n\), so set \(b_n=1/n\). Both \(a_n\) and \(b_n\) are positive for every \(n\geq1\).
Step 2: Compute the ratio and its limit.
Divide the numerator and denominator by \(n^3\):
The limit is positive and finite, so limit comparison applies. The harmonic series \(\sum 1/n\) diverges; therefore \(\sum a_n\) diverges as well.
Conclusion and check: The given series does not converge absolutely. The ratio limit is \(3\), satisfying \(0<3<\infty\), and the benchmark diverges. This comparison does not determine whether the original signed series converges; it establishes only that absolute convergence fails.
Worked Example: A Power-Type Series
Determine whether
converges absolutely.
Step 1: Form the absolute-value terms. The positive terms to test are
Compare with \(b_n=1/n^{3/2}\), whose p-series converges because \(3/2>1\).
Step 2: Compute the comparison ratio.
Dividing numerator and denominator by \(n^{7/2}\) gives
Since the ratio limit is \(1\), which is positive and finite, limit comparison shows that \(\sum a_n\) has the same behavior as \(\sum 1/n^{3/2}\). The latter converges.
Conclusion and check: The given series converges absolutely and therefore converges. The chosen benchmark has exponent \(3/2>1\), and the computed ratio tends to \(1\), so both requirements for this comparison are satisfied.
Choosing and Checking a Benchmark
For rational terms, the highest powers often suggest a useful benchmark. If the absolute value of a rational term behaves like a constant times \(1/n^p\), compare it with \(\sum 1/n^p\). When \(p>1\), that comparison can establish absolute convergence; when \(p\leq1\), a positive finite limit comparison with that p-series establishes divergence of the absolute-value series.
The benchmark must match the behavior closely enough for the chosen test. With direct comparison, prove an inequality in the correct direction. With limit comparison, compute the ratio and verify that its limit is strictly positive and finite. Do not infer a conclusion from a ratio limit of zero or infinity using the finite-positive-limit theorem.
A reliable routine is therefore: remove the signs, identify the scale of the positive terms, select a benchmark with known behavior, and verify the hypotheses of the comparison theorem before stating a conclusion.
Check Your Understanding
For each question, focus on the absolute-value series and on the conditions needed for comparison.
- What does it mean for \(\sum u_n\) to converge absolutely?
- Suppose \(0\leq |u_n|\leq 4/n^3\) for all sufficiently large \(n\). What can you conclude, and which comparison result justifies it?
- If \(a_n>0\), \(b_n>0\), and \(a_n/b_n\to 5\), what does limit comparison say about \(\sum a_n\) and \(\sum b_n\)?
- Why does divergence of \(\sum |u_n|\) fail to decide whether \(\sum u_n\) converges?
- For an absolute-value term whose ratio to \(1/n\) tends to \(2\), what does limit comparison establish about absolute convergence?
