Why Compare the Magnitudes?
An alternating series may converge because its positive and negative terms cancel. But cancellation is not the only way a series can converge. If the sum of the term magnitudes is finite, then the original series converges regardless of its signs. This stronger property is called absolute convergence.
For an alternating series, the key question is therefore often whether the positive-term series formed from the magnitudes converges. Direct and limit comparison answer that question by relating its terms to a familiar series, often a p-series.
The signs of \(u_n\) do not appear in the convergence tests for \(\sum |u_n|\): its terms are nonnegative. In the examples below, the original terms alternate, but every comparison is made using their magnitudes.
Direct Comparison
The direct comparison test applies to nonnegative terms. If \(0\le a_n\le b_n\) for all \(n\) from some index onward, then convergence of \(\sum b_n\) implies convergence of \(\sum a_n\). The inequality only needs to hold eventually: changing or omitting finitely many terms does not change whether a series converges.
A common reference series is the p-series \(\sum 1/n^p\), which converges for \(p>1\) and diverges for \(p\le 1\). For absolute convergence, the useful direction is usually to bound the magnitudes above by the terms of a convergent p-series.
Worked Example: A Rational Expression with Polynomial Terms
Determine whether
converges absolutely. Its magnitude term is
because the numerator and denominator are positive for \(n\ge 1\). For these indices, \(3n+2\le 5n\) and \(n^3+1\ge n^3\). Therefore
The comparison series \(\sum 5/n^2\) converges because it is a constant multiple of a p-series with \(p=2>1\). Direct comparison gives \(\sum a_n<\infty\), so the original alternating series converges absolutely. The inequalities hold starting at \(n=1\), and the final comparison is exactly with the convergent p-series term \(5/n^2\).
Polynomial and Logarithmic Factors
Logarithms grow more slowly than every positive power of their input. In particular, for \(x\ge 1\), \(\ln(x+1)\le \sqrt{x}\). One way to verify this bound is to set \(g(x)=\sqrt{x}-\ln(x+1)\). Then
Also \(g(1)=1-\ln 2>0\). Thus \(g(x)>0\) for \(x\ge1\), giving the stated inequality. This type of bound lets us replace a logarithmic factor with a power that is convenient for comparison.
Worked Example: A Logarithm over a Quadratic
Determine whether
converges absolutely. Its magnitude term is \(a_n=\ln(n+1)/n^2\), which is nonnegative. Applying \(\ln(n+1)\le\sqrt n\) gives
The series \(\sum 1/n^{3/2}\) converges because \(3/2>1\). By direct comparison, \(\sum \ln(n+1)/n^2\) converges, so the alternating series converges absolutely. For instance, at \(n=1\) the bound reads \(\ln 2\le1\), consistent with the general inequality.
Limit Comparison
Sometimes a clean inequality is difficult to find, but the ratio of two positive terms has a simple limit. The limit comparison test then transfers convergence or divergence between the two series.
Here is why the test works. Since \(a_n/b_n\to c>0\), for all sufficiently large \(n\) the ratio lies between \(c/2\) and \(3c/2\). Multiplying by \(b_n>0\) gives
The terms are therefore bounded above and below by positive constant multiples of one another from some index onward. Direct comparison in both directions shows that the two series have the same convergence behavior.
Worked Example: Matching Polynomial and Logarithmic Factors
Determine whether
converges absolutely. Let \(a_n\) be the magnitude of its \(n\)th term, and compare it with \(b_n=\ln(n+1)/n^2\). The comparison series \(\sum b_n\) converges by the preceding example. For every \(n\ge1\), both \(a_n\) and \(b_n\) are positive, so
Dividing numerator and denominator by \(n\) shows the limit:
The limit is finite and positive, so the limit comparison test applies. Since \(\sum b_n\) converges, \(\sum a_n\) also converges. Thus the original series converges absolutely. The computed ratio simplifies exactly to \((2n+1)/(n+5)\), whose numerator and denominator have leading coefficients \(2\) and \(1\), respectively, confirming the limit \(2\).
When the Ratio Limit Is Zero
The finite-positive-limit condition has a useful one-way extension. If \(a_n,b_n>0\), \(a_n/b_n\to0\), and \(\sum b_n\) converges, then \(\sum a_n\) converges. Indeed, eventually \(a_n/b_n<1\), so \(a_n<b_n\), and direct comparison applies. This is especially useful when a logarithmic factor is outweighed by a larger power in the denominator.
Worked Example: A Logarithm Outweighed by a Power
Determine whether
converges absolutely. Compare its magnitude term \(a_n=\ln(n+1)/n^{5/2}\) with \(b_n=1/n^2\). The reference series converges since \(p=2>1\). The ratio is
The bound \(\ln(n+1)\le\sqrt n\) alone only bounds this ratio by \(1\), so use the same growth fact with \(x=\sqrt n\): \(\ln(\sqrt n+1)\le n^{1/4}\). Since \(n+1\le(\sqrt n+1)^2\), taking logarithms gives \(\ln(n+1)\le2\ln(\sqrt n+1)\le2n^{1/4}\). Consequently,
Because the ratio tends to zero and \(\sum 1/n^2\) converges, the zero-limit comparison gives convergence of \(\sum a_n\). The alternating series therefore converges absolutely. In particular, the estimate supplies the needed limit rather than relying only on the fact that a logarithm grows more slowly than a positive power.
A Common Comparison Mistake
A ratio limit of zero does not allow an unrestricted conclusion: the direction of comparison and the behavior of the reference series matter. For example, take \(a_n=1/n^2\) and \(b_n=1/n\). Then
It would be wrong to say that \(\sum a_n\) diverges because the harmonic series \(\sum b_n\) diverges. In fact, \(\sum a_n=\sum 1/n^2\) converges. A zero ratio gives a convergence conclusion when the comparison series converges; when the comparison series diverges, it gives no such conclusion. The direct comparison test likewise requires the inequality in the correct direction: an upper bound by a divergent series does not prove convergence.
Why These Comparisons Matter
An alternating-series test can show convergence without showing absolute convergence. Comparing magnitudes separates those two questions: the comparison tests decide whether the absolute-value series converges, while any conclusion about the original signed series follows from that result. This is particularly effective when a term contains both polynomial and logarithmic factors, because polynomial powers often determine a useful p-series comparison and logarithms can be bounded or retained in a limit comparison.
A reliable solution makes each link explicit: define the nonnegative magnitude terms, name the comparison series, verify the required bound or limit, state why the reference series converges, and then conclude absolute convergence. If any one of those links is missing, the comparison may not justify the claim.
Check Your Understanding
For each question, identify the positive series used for comparison and state which test applies.
- What inequality would you seek to prove to show that \(\sum (-1)^{n-1}(n+1)/(n^3+2)\) converges absolutely by direct comparison with a p-series?
- If \(a_n,b_n>0\) eventually and \(a_n/b_n\to 4\), what does the limit comparison test say about the convergence of their series?
- Why does \(a_n/b_n\to0\), together with convergence of \(\sum b_n\), imply convergence of \(\sum a_n\)?
- For \(a_n=1/n^2\) and \(b_n=1/n\), what is the ratio limit, and why does the divergence of \(\sum b_n\) not imply divergence of \(\sum a_n\)?
- When testing absolute convergence of an alternating series, which terms should be used in a comparison: the signed terms or their magnitudes?
