Why Logarithmic Series Need Careful Comparisons
A logarithmic factor can change whether a series converges when the power of \(n\) is exactly at a critical value. But when the power of \(n\) is away from that value, it is usually the decisive factor: fixed powers of logarithms grow more slowly than every positive power of \(n\).
The practical task is to identify which factor controls the series, then compare its terms with a model whose convergence behavior is known. The limit comparison test is useful when the ratio of the two terms approaches a positive finite number. If the ratio approaches zero or infinity, a different argument is needed.
The Model Thresholds
For \(n\ge 3\), consider the model terms
where \(p,q,r\) are real numbers. For these terms, the first exponent to check is \(p\). If \(p\ne 1\), the power of \(n\) decides the classification, regardless of the fixed logarithmic exponents. If \(p=1\), check \(q\); if \(q=1\) as well, check \(r\).
This is a hierarchy of thresholds. A logarithmic exponent matters only after the preceding exponent reaches its critical value. For example, when \(p=1\) and \(q=1\), the remaining threshold is the iterated-logarithm exponent \(r=1\).
The same pattern extends to further iterated logarithms. Write \(\log_1 n=\log n\), \(\log_2 n=\log\log n\), and recursively \(\log_{j+1}n=\log(\log_j n)\), considering \(n\) large enough that the expressions are positive. In a model with denominator \(n\log_1 n\cdots\log_k n\), a further factor \((\log_{k+1}n)^\alpha\) converges when \(\alpha>1\) and diverges when \(\alpha\le 1\). Only finitely many initial terms need to be excluded to make these factors positive.
Limit Comparison: The Exact Condition
Let \(a_n>0\) and \(b_n>0\) for all sufficiently large \(n\). If
then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge.
Here is why. Since the ratio tends to \(L>0\), it eventually lies between \(L/2\) and \(3L/2\). Thus, for all sufficiently large \(n\),
If \(\sum b_n\) converges, the upper bound and the comparison test imply that \(\sum a_n\) converges. If \(\sum a_n\) converges, the lower bound gives \(b_n\le (2/L)a_n\), so the comparison test implies that \(\sum b_n\) converges. These implications establish the result in both directions.
The requirement \(0<L<\infty\) matters. A ratio tending to zero or infinity is not the conclusion of this theorem. Such a limit can still be useful: it may establish an eventual inequality for the direct comparison test.
Worked Comparisons
Worked Example: A Power Above the Critical Value
Classify
Set \(a_n=(1+2/\log n)/n^{3/2}\), and compare with \(b_n=1/n^{3/2}\). Both terms are positive for \(n\ge3\). The ratio is
The limit is positive and finite, so limit comparison applies. The comparison series is a \(p\)-series with exponent \(3/2>1\), and therefore converges. Hence the given series converges. The logarithmic factor changes the ratio by a factor tending to \(1\), so it does not change this classification.
Check: the computed ratio has limit \(1\), not zero or infinity; the model exponent is \(3/2\), which is greater than the convergence threshold \(1\). Both checks agree with the conclusion.
Worked Example: A Logarithm Below Its Critical Exponent
Classify
Here the power of \(n\) is at its threshold \(p=1\), but the exponent of \(\log n\) is \(q=1/2<1\). The model classification predicts divergence, and we can verify it by direct comparison. Choose
This comparison series diverges because its \(n\)-exponent is \(1\) and its logarithmic exponent is \(3/4<1\). The ratio is
The power \((\log n)^{1/4}\) eventually exceeds \((\log\log n)^2\), since a positive power of \(\log n\) dominates any fixed power of \(\log\log n\). Thus \(a_n/b_n\ge1\) for all sufficiently large \(n\), so \(a_n\ge b_n\) there. Since the smaller comparison series diverges, the given series diverges as well.
Check: the ratio tends to infinity, giving the required eventual lower bound \(a_n\ge b_n\); the comparison series diverges because \(3/4<1\). The direct comparison therefore confirms the classification.
Worked Example: The Iterated-Logarithm Threshold
Classify
Compare with \(b_n=1/[n\log n(\log\log n)^2]\). The ratio is
The model series \(b_n\) has \(p=q=1\) and \(r=2>1\), so it converges by the iterated-logarithm threshold. The ratio has a positive finite limit, so limit comparison shows that the given series converges too.
Check: the extra factor in the numerator tends to \(1\), and the model’s last exponent is \(2\), above its threshold \(1\). Both facts support convergence.
Worked Example: A Logarithm at the Divergence Threshold
Classify
Rewrite the term so its leading logarithmic behavior is visible:
Let \(b_n=1/(n\log n)\). Then
The series with terms \(b_n\) is the harmonic logarithmic threshold and diverges. Since the ratio has a positive finite limit, the given series also diverges.
Check: rewriting the numerator produces the factor \(1+4/\log n\), which tends to \(1\), and the remaining model has logarithmic exponent \(q=1\). That is the divergent threshold case.
When the Power of \(n\) Decides
A frequent mistake is to focus on a large logarithmic factor while overlooking that the power of \(n\) is already above its critical value. For example, \((\log n)^5/n^2\) has a growing logarithmic numerator, but the series still converges. For sufficiently large \(n\), the power-dominance property gives \((\log n)^5\le n^{1/2}\). Therefore,
The series on the right converges, so direct comparison proves convergence on the left. The logarithm grows, but not quickly enough to overcome the power \(n^2\).
The first series converges, while the second diverges. The ratio limit is zero, so the positive-finite-limit version of limit comparison does not apply. The correction is to use the appropriate logarithmic threshold for \(a_n\), or to use a valid eventual inequality; do not infer matching behavior from a zero ratio.
A Reliable Decision Process
For a positive series involving powers and logarithms, organize the comparison before calculating. First identify the exponent of \(n\). If it is not \(1\), that power determines the model classification. If it is \(1\), inspect the exponent of \(\log n\); if that exponent is also \(1\), inspect the exponent of \(\log\log n\). Then choose a model series whose behavior is known.
Ensure the logarithms and all terms are positive for sufficiently large \(n\). Finitely many initial terms do not affect convergence.
Check the power of \(n\), then the logarithm, then iterated logarithms if earlier exponents equal their thresholds.
Use a model with the same leading behavior when possible. Compute the ratio rather than relying only on visual resemblance.
A positive finite ratio limit permits limit comparison. A zero or infinite ratio may support direct comparison, but does not by itself establish matching convergence behavior.
The essential habit is to distinguish the critical exponent from factors that merely modify the terms by a slower growth scale. Once the first threshold is identified, a comparison ratio can confirm whether the expression has the same convergence behavior as its model.
Check Your Understanding
For each question, identify the relevant threshold or comparison principle before deciding on convergence.
- Does \(\sum_{n=3}^{\infty}1/[n^{5/4}(\log n)^7]\) converge or diverge? Which exponent decides?
- Classify \(\sum_{n=3}^{\infty}1/[n(\log n)^{6/5}(\log\log n)^4]\). Does the iterated-logarithm exponent affect the decision?
- For \(a_n=1/[n\log n(\log\log n)^{3/2}]\), which model exponents determine the classification?
- If \(a_n/b_n\to0\), may limit comparison alone establish that \(\sum a_n\) and \(\sum b_n\) have the same behavior? Explain.
- Find the limit of the ratio between \((1+3/\log n)/(n\log n)\) and \(1/(n\log n)\), and state what that limit permits you to conclude.
