Why Parameter Thresholds Matter
In many series, changing a single exponent changes the conclusion from convergence to divergence. The important task is not just to test one numerical case, but to locate the exact threshold and check what happens at the boundary.
For series involving powers of \(n\), the \(p\)-series \(\sum 1/n^p\) is a basic comparison benchmark: it converges exactly when \(p>1\). Logarithmic factors can change the outcome at the boundary \(p=1\), while alternating signs can allow convergence even when the series of absolute values diverges.
The Positive Power-Log Family
Consider the positive-term series
where \(p\) and \(q\) are fixed real numbers. Starting at \(n=2\) ensures that \(\ln n\) is positive, so real powers of \(\ln n\) are defined.
The power of \(n\) determines the result except at \(p=1\). At that boundary, the logarithmic exponent supplies the threshold.
Why Those Thresholds Hold
The comparison idea behind the theorem is that every fixed power of \(\ln n\) grows more slowly than every positive power of \(n\). In particular, for \(r>0\) and \(\varepsilon>0\),
This limit allows a logarithmic factor to be bounded above or below by a small change in the power of \(n\). The \(p\)-series comparison then settles the cases \(p\neq 1\). When \(p=1\), the logarithm itself determines convergence.
If \(q\geq 0\), then \((\ln n)^{-q}\leq 1\) for \(n\geq 2\), so the terms are at most \(1/n^p\), a convergent \(p\)-series. If \(q<0\), write \(r=-q>0\) and choose \(0<\varepsilon<p-1\). Eventually \((\ln n)^r\leq n^\varepsilon\), so the terms are at most \(1/n^{p-\varepsilon}\), also a convergent \(p\)-series.
If \(q\leq 0\), the terms are at least \(1/n^p\), whose series diverges. If \(q>0\), choose \(0<\varepsilon<1-p\). Eventually \((\ln n)^q\leq n^\varepsilon\), so the terms are at least \(1/n^{p+\varepsilon}\). Since \(p+\varepsilon<1\), this comparison series diverges.
The integral test for logarithmic series gives convergence of \(\sum 1/[n(\ln n)^q]\) exactly when \(q>1\). Indeed, the corresponding integral becomes \(\int u^{-q}\,du\) after the substitution \(u=\ln x\); its tail is finite exactly when \(q>1\).
This proof explains why the dividing line is not simply “large denominator means convergence.” For \(p<1\), even a positive logarithmic power cannot compensate for the insufficient power of \(n\). At \(p=1\), however, the logarithm decides the outcome.
Worked Example: A Power Below the Threshold
Worked Example: Classify a Power-Log Series with \(p<1\)
Determine whether \(\sum_{n=2}^{\infty}1/[n^{3/4}(\ln n)^{7/2}]\) converges.
Here \(p=3/4\) and \(q=7/2\). Since \(p<1\), the theorem predicts divergence, but we can see the comparison explicitly. Choose \(\varepsilon=1/8\). Then \(p+\varepsilon=3/4+1/8=7/8<1\). Because \((\ln n)^{7/2}/n^{1/8}\to 0\), eventually \((\ln n)^{7/2}\leq n^{1/8}\). Therefore, for all sufficiently large \(n\),
The series \(\sum 1/n^{7/8}\) diverges because \(7/8\leq 1\). The given terms are at least as large eventually, so the given series diverges by comparison. The comparison checks out: its lower-bound series is a divergent \(p\)-series, as required.
The Logarithmic Boundary in Detail
At \(p=1\), the family becomes \(\sum 1/[n(\ln n)^q]\). The integral-test comparison is valid because \(f(x)=1/[x(\ln x)^q]\) is positive and eventually decreasing for every fixed real \(q\). To see the boundary directly, substitute \(u=\ln x\), so \(du=dx/x\):
The last integral converges for \(q>1\) and diverges for \(q\leq 1\). This includes negative \(q\): in that case the integrand is \(u^{|q|}\), which certainly does not have a finite integral over an unbounded interval.
Worked Example: Test the Logarithmic Boundary
Classify \(\sum_{n=2}^{\infty}1/[n(\ln n)^{3/2}]\).
This is the boundary case \(p=1\) with \(q=3/2\). The integral-test integral is
This value is finite, so the series converges by the integral test. The result agrees with the threshold \(q>1\). As a boundary check, changing \(q\) to \(1\) gives \(\int du/u\), which diverges, so the series with terms \(1/[n\ln n]\) diverges.
Alternating Power-Log Series
Now consider the alternating family
For alternating-series convergence, a standard sufficient test requires the positive magnitudes to decrease eventually and approach zero. If both conditions hold, the alternating series converges. Absolute convergence is determined separately by applying the positive power-log classification to the magnitudes.
For \(b(x)=x^{-p}(\ln x)^{-q}\), its derivative has the same sign as the logarithmic derivative
If \(p>0\), then \(p+q/\ln x>0\) for all sufficiently large \(x\), whatever fixed real value \(q\) has. Thus the magnitudes eventually decrease. When \(p=0\) and \(q>0\), the expression is also positive, so the magnitudes decrease in that case as well.
Combining the zero-limit condition, the alternating-series test, and the absolute-convergence threshold gives the complete classification.
For the conditional cases with \(p>0\), the magnitudes approach zero because a positive power of \(n\) dominates any fixed logarithmic power. For \(p=0,q>0\), the magnitudes are \(1/(\ln n)^q\), which also approach zero. The decreasing condition follows from the derivative calculation. In these conditional cases, the positive series of magnitudes diverges by the power-log threshold, so convergence is not absolute.
Worked Example: Conditional Convergence at \(p=1\)
Worked Example: Separate Alternating and Absolute Convergence
Classify \(\sum_{n=2}^{\infty}(-1)^n/[n(\ln n)^{1/2}]\).
The magnitude is \(b_n=1/[n(\ln n)^{1/2}]\), corresponding to \(p=1\) and \(q=1/2\). To verify eventual decrease, use \(b(x)=x^{-1}(\ln x)^{-1/2}\). The derivative sign is determined by
Also, \(b_n\to 0\), since its denominator grows without bound. The alternating-series test therefore proves that the series converges. For absolute convergence, the magnitudes form the positive series \(\sum 1/[n(\ln n)^{1/2}]\). Here \(p=1\) and \(q=1/2\leq 1\), so the logarithmic boundary test says this positive series diverges. Thus the alternating series is conditionally convergent. Both checks are necessary: the alternating test proves convergence, while the positive-series test rules out absolute convergence.
Worked Example: A Zero-Power Alternating Boundary
Worked Example: Alternation with a Logarithmic Decay
Classify \(\sum_{n=2}^{\infty}(-1)^n/\ln n\).
This is the alternating family with \(p=0\) and \(q=1\). Its magnitudes are \(b_n=1/\ln n\), which tend to zero. For \(x>1\),
so the magnitudes decrease. The alternating-series test proves convergence. The absolute-value series \(\sum 1/\ln n\) diverges: for \(n\geq 2\), \(\ln n<n\), so \(1/\ln n>1/n\), and the harmonic series diverges. Therefore the alternating series converges conditionally. The comparison confirms that its absolute values do not converge.
A Common Mistake: Treating Zero Limit as Enough
A common error is to conclude that a series converges merely because its terms approach zero. For example, the terms \(1/[n\ln n]\) do approach zero, but the series \(\sum 1/[n\ln n]\) diverges: it is the \(p=1,q=1\) boundary, where the logarithmic test requires \(q>1\).
A related mistake is to treat an alternating series and its absolute-value series as if they had the same convergence result. For \(\sum (-1)^n/[n(\ln n)^{1/2}]\), the alternating-series test applies, but the absolute-value series diverges. Its convergence is conditional, not absolute. Always check the terms’ limit, the applicable test’s other hypotheses, and absolute convergence as separate questions.
Using Thresholds Efficiently
For a family with parameters, first rewrite the term so the power of \(n\) and the logarithmic power are visible. Then locate the parameter boundary before substituting particular values. The main benchmarks are:
- For positive power-log terms, compare the power of \(n\) with the \(p=1\) threshold.
- If the power is exactly \(1\), compare the logarithmic exponent with \(q=1\).
- For alternating terms, check absolute convergence first, then verify eventual decrease and a zero limit if absolute convergence fails.
- If the terms do not tend to zero, stop: the series diverges, regardless of its signs.
This approach prevents a boundary case from being decided by intuition alone. It also makes clear which standard test supports each conclusion and what that test requires.
Check Your Understanding
For each question, state the relevant parameter values and the test or comparison that settles the conclusion.
- Does \(\sum_{n=2}^{\infty}1/[n^{5/4}(\ln n)^{-2}]\) converge?
- For which values of \(q\) does \(\sum_{n=2}^{\infty}1/[n(\ln n)^q]\) converge?
- Classify \(\sum_{n=2}^{\infty}(-1)^n/[n^{2/3}(\ln n)^4]\) as absolutely convergent, conditionally convergent, or divergent.
- Does \(\sum_{n=2}^{\infty}(-1)^n/(\ln n)^{-1}\) satisfy the necessary condition that its terms approach zero?
- Why does convergence of an alternating series not by itself establish absolute convergence?
