Why Compare the Power and Logarithm Separately?
A series term can contain both a power of the index and a power of its logarithm. For instance, \(\frac{(\ln n)^4}{n^{3/2}}\) has a logarithmic factor that increases as \(n\) increases, while its power of \(n\) decreases the term. To decide whether the series converges, we need to determine which factor controls the terms for large \(n\).
When the exponent of \(n\) is not one, a power-series benchmark gives a decisive comparison. The logarithmic factor may increase or decrease, but it does not overturn the convergence or divergence dictated by the power of \(n\). We will establish this using limits and then apply it to specific series.
Power-Logarithm Terms
The series under consideration is \(\sum_{n=2}^{\infty} n^{-p}(\ln n)^q\). The exponents \(p\) and \(q\) may be any real numbers. In particular, a negative \(q\) means that the logarithmic factor is in the denominator: \((\ln n)^q=1/(\ln n)^{|q|}\).
We will use the p-series criterion: \(\sum_{n=1}^{\infty}n^{-r}\) converges if \(r>1\) and diverges if \(r\le 1\). The comparison and limit-comparison tests for positive terms let us compare a series with one of these benchmarks.
The Limit That Makes the Comparison Work
The key fact is that a fixed power of a logarithm is smaller, in the limit, than any positive power of \(n\). Here is the particular limit we need, including a proof.
To prove this, set \(x=\ln n\). As \(n\to\infty\), \(x\to\infty\), and \(n^\delta=e^{\delta x}\). It is enough to show that \(x^s e^{-\delta x}\to 0\). If \(s\le 0\), then \(x^s\le 1\) for \(x\ge 1\), so \(0\le x^s e^{-\delta x}\le e^{-\delta x}\), which tends to zero.
If \(s>0\), choose an integer \(m>s\). The exponential series gives \(e^{\delta x}\ge(\delta x)^m/m!\) for \(x>0\). Therefore,
Because \(s-m<0\), the right-hand side tends to zero as \(x\to\infty\). This proves the limit for every real \(s\). Taking reciprocals also shows that \(n^\delta/(\ln n)^s\to\infty\).
Classification Away from the Exponent One
The proof uses a power-series benchmark with exponent between \(p\) and \(1\). Such a choice is possible precisely because \(p\ne 1\).
Choose \(r\) so that \(1<r<p\). Then \(n^{-r}\) is a convergent p-series.
The ratio to the benchmark is \(n^{r-p}(\ln n)^q=(\ln n)^q/n^{p-r}\), which tends to zero by the limit just proved.
Choose \(r\) so that \(p<r<1\). Then \(n^{-r}\) is a divergent p-series, and the ratio \(n^{r-p}(\ln n)^q\) tends to infinity.
To complete the reasoning, consider each case. If \(p>1\), the ratio \(a_n/n^{-r}\) tends to zero. It is therefore less than \(1\) for all sufficiently large \(n\), so \(a_n<n^{-r}\) eventually. Comparison with the convergent series \(\sum n^{-r}\) proves convergence.
If \(p<1\), set \(\delta=r-p>0\). The reciprocal of the ratio is
Thus \(a_n/n^{-r}\to\infty\), so \(a_n>n^{-r}\) for all sufficiently large \(n\). Comparison with the divergent series \(\sum n^{-r}\) proves divergence. Changing or omitting finitely many initial terms does not affect convergence, so starting the series at \(n=2\) causes no difficulty.
Worked Examples
Worked Example: A Growing Logarithmic Factor
Classify \(\sum_{n=2}^{\infty}\frac{(\ln n)^2}{n^{3/2}}\). Here \(p=\frac32>1\) and \(q=2\), so the theorem predicts convergence. To verify it with a limit comparison, choose the benchmark exponent \(r=\frac54\), which satisfies \(1<\frac54<\frac32\). The benchmark series \(\sum n^{-5/4}\) converges.
The limit follows from the proved result with \(s=2\) and \(\delta=\frac14\). Consequently, the given terms are eventually smaller than \(n^{-5/4}\), and the series converges by comparison. The ratio tends to zero, as required for this comparison with a convergent benchmark.
Worked Example: A Negative Logarithmic Power
Classify \(\sum_{n=2}^{\infty}\frac{1}{n^{2/3}(\ln n)^2}\). Its exponents are \(p=\frac23<1\) and \(q=-2\). Choose \(r=\frac56\), so \(\frac23<\frac56<1\). The benchmark \(\sum n^{-5/6}\) diverges.
The reciprocal of this ratio is \((\ln n)^2/n^{1/6}\), which tends to zero by the limit with \(s=2\) and \(\delta=\frac16\). Hence the displayed ratio tends to infinity. The given terms are eventually larger than \(n^{-5/6}\), so comparison with the divergent benchmark proves that the given series diverges. In particular, the decreasing logarithmic factor does not make the series converge.
Worked Example: A Large Positive Logarithmic Power
Classify \(\sum_{n=2}^{\infty}\frac{(\ln n)^{50}}{n^{9/10}}\). Although the logarithmic factor grows, the exponent \(p=\frac9{10}\) is less than one. Choose \(r=\frac{19}{20}\), which lies between \(\frac9{10}\) and \(1\). The benchmark \(\sum n^{-19/20}\) diverges.
The reciprocal is \((\ln n)^{-50}/n^{1/20}\), which tends to zero by the proved limit with \(s=-50\) and \(\delta=\frac1{20}\). Thus the ratio tends to infinity, and the given terms eventually exceed the benchmark terms. The series diverges by comparison. The limit, rather than the apparent size of the logarithmic factor at small indices, establishes the conclusion.
Common Pitfall: Treating an Unbounded Factor as Decisive
A common error is to notice that \((\ln n)^{50}\) grows without bound and conclude that it must overturn the decay of a power of \(n\). For example, one might incorrectly reason that \(\sum (\ln n)^{50}/n^{9/10}\) converges because the denominator contains a power of \(n\). Neither factor should be considered in isolation.
The correction is to compare with a benchmark whose exponent is strictly between \(\frac9{10}\) and \(1\), such as \(\frac{19}{20}\). The ratio is \(n^{1/20}(\ln n)^{50}\), and its reciprocal tends to zero. Therefore the terms are eventually larger than those of a divergent p-series, so the series diverges. The limit resolves what informal impressions about “large” and “small” factors cannot.
Why the Exponent One Is Different
The comparison relies on choosing a benchmark exponent strictly between \(p\) and \(1\). If \(p>1\), that benchmark must also be greater than \(1\); if \(p<1\), it must also be less than \(1\). When \(p=1\), there is no exponent strictly between \(p\) and \(1\), so this argument does not decide the result.
This boundary case needs a separate comparison that accounts more finely for the logarithmic power. The conclusion established here is limited to \(p\ne1\): on either side of one, the power exponent determines convergence or divergence, regardless of the fixed logarithmic exponent.
Check Your Understanding
For each question, use a power-series benchmark and state the limit that supports your comparison.
- Does \(\sum_{n=2}^{\infty}(\ln n)^3/n^{7/5}\) converge or diverge?
- Does \(\sum_{n=2}^{\infty}1/[n^{3/4}(\ln n)^4]\) converge or diverge?
- For \(p>1\), what conditions must a benchmark exponent \(r\) satisfy?
- Why does \((\ln n)^s/n^\delta\to0\) for every real \(s\) when \(\delta>0\)?
- Why does the benchmark argument in this tutorial not settle the case \(p=1\)?