The Harmonic Logarithmic Threshold

Why the Logarithmic Factor Has a Threshold

The harmonic series \(\sum 1/n\) diverges, but adding a factor of \(\log n\) to the denominator makes its terms smaller. How much of that factor is needed to make the series converge? The answer depends on the power of the logarithm, and there is a precise boundary between convergence and divergence.

Throughout this tutorial, \(\log\) denotes the natural logarithm. We start sums at \(n=2\), so \(\log n>0\); this also makes \((\log n)^p\) well-defined and positive for every real exponent \(p\).

Threshold theorem: For real \(p\), the positive-term series \[ \sum_{n=2}^{\infty}\frac{1}{n(\log n)^p} \] converges if and only if \(p>1\). It diverges for \(p\leq 1\). The exponent \(p=1\) is the critical case.

This threshold gives a comparison model. If another positive-term series has terms asymptotic to a positive constant times \(1/[n(\log n)^p]\), limit comparison transfers the same convergence or divergence behavior to that series.

Deriving the Threshold

We establish the classification using the integral test. The test applies to a series \(\sum f(n)\) when \(f\) is positive, continuous, and decreasing for all sufficiently large \(x\). Under those conditions, the series and the improper integral \(\int f(x)\,dx\) have the same convergence behavior. Changing or removing finitely many initial terms does not affect convergence.

Set \(f(x)=1/[x(\log x)^p]\) for \(x>1\). It is positive and continuous. To check that it is eventually decreasing, differentiate:

f'(x) = -\frac{1}{x^2(\log x)^p} \left(1+\frac{p}{\log x}\right).

For any fixed real \(p\), the factor \(1+p/\log x\) is positive once \(x\) is sufficiently large. Thus \(f'(x)<0\) eventually, and the integral test applies.

To evaluate the improper integral, use \(u=\log x\), so \(du=dx/x\). The convergence question becomes the convergence of \(\int_{\log 2}^{\infty}u^{-p}\,du\). If \(p\neq 1\), an antiderivative is \(u^{1-p}/(1-p)\); if \(p=1\), an antiderivative is \(\log u\).

\int_2^\infty \frac{dx}{x(\log x)^p} = \begin{cases} \displaystyle \int_{\log 2}^{\infty}u^{-p}\,du = \left[\frac{u^{1-p}}{1-p}\right]_{\log 2}^{\infty}, & p\neq 1,\\[1.2em] \displaystyle \int_{\log 2}^{\infty}\frac{du}{u} = \left[\log u\right]_{\log 2}^{\infty}, & p=1. \end{cases}

When \(p>1\), \(1-p<0\), so \(u^{1-p}\) tends to zero as \(u\) tends to infinity. The integral is finite. When \(p<1\), \(1-p>0\), so \(u^{1-p}\) grows without bound and the integral diverges. At \(p=1\), \(\log u\) grows without bound, so that integral also diverges. The integral test now gives the stated series classification.

Why the boundary is \(1\): The substitution \(u=\log x\) turns the factor \(dx/x\) into \(du\). What remains is the power integral \(\int u^{-p}\,du\), whose convergence changes at \(p=1\).

Using the Threshold with Limit Comparison

The threshold theorem classifies the standard model. To classify a different positive-term series \(\sum a_n\), choose a model \(b_n=1/[n(\log n)^p]\) with the same leading behavior. If

\lim_{n\to\infty}\frac{a_n}{b_n}=c \qquad\text{for some }0<c<\infty,

then the limit comparison test says that \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. In practice, the ratio often simplifies to a familiar expression whose limit can be evaluated directly.

The requirement that \(c\) be finite and strictly positive matters. A zero or infinite limit does not, by itself, allow the same conclusion. We will return to that distinction after the examples.

Worked Example: A Convergent Series with Logarithmic Power Two

Classify \[ \sum_{n=2}^{\infty}\frac{3n+1}{n^2(\log n)^2}. \]

Every term is positive for \(n\geq 2\). Compare with \(b_n=1/[n(\log n)^2]\). The ratio is

\frac{a_n}{b_n} = \frac{3n+1}{n^2(\log n)^2} \cdot n(\log n)^2 = \frac{3n+1}{n} = 3+\frac{1}{n}.

Therefore \(\lim_{n\to\infty}a_n/b_n=3\), which is finite and positive. The model has exponent \(p=2>1\), so \(\sum b_n\) converges by the threshold theorem. Limit comparison implies that the given series converges as well.

Check: the ratio tends to \(3\), not to zero or infinity, and the comparison model is on the convergent side of the threshold. Both conditions needed for this conclusion are satisfied.

Worked Example: The Critical Exponent

Classify \[ \sum_{n=2}^{\infty}\frac{\sqrt{n^2+4n+1}}{n^2\log n}. \]

The terms are positive. Choose \(b_n=1/[n\log n]\), the model at \(p=1\). Compute the ratio:

\frac{a_n}{b_n} = \frac{\sqrt{n^2+4n+1}}{n^2\log n} \cdot n\log n = \frac{\sqrt{n^2+4n+1}}{n} = \sqrt{1+\frac{4}{n}+\frac{1}{n^2}}.

As \(n\to\infty\), the terms inside the square root tend to \(1\), so the ratio tends to \(1\). This is a finite positive limit. The model series has \(p=1\), so it diverges by the critical case of the threshold theorem. Limit comparison implies that the given series diverges.

Check: factoring \(n^2\) from beneath the square root gives the last expression exactly, and its limit is \(1\). Thus the comparison preserves the model's divergence.

Worked Example: A Logarithmic Power Below the Threshold

Classify \[ \sum_{n=2}^{\infty}\frac{2n+5}{n^2\sqrt{\log n}}. \]

Here \(\sqrt{\log n}=(\log n)^{1/2}\), so compare with \(b_n=1/[n(\log n)^{1/2}]\). The terms are positive, and

\frac{a_n}{b_n} = \frac{2n+5}{n^2(\log n)^{1/2}} \cdot n(\log n)^{1/2} = \frac{2n+5}{n} = 2+\frac{5}{n}.

The ratio tends to \(2\), a finite positive number. The comparison model has \(p=1/2\leq 1\), so it diverges. By limit comparison, the given series diverges.

Check: \((\log n)^{1/2}\) corresponds to exponent \(p=1/2\), not \(p=2\). Since that exponent lies below the critical value, divergence is the correct conclusion.

When a Comparison Ratio Is Not Enough

A frequent error is to treat a ratio tending to zero as if it always proved convergence. It does not. For positive terms, a zero limit only says that \(a_n\) is eventually smaller than a constant multiple of \(b_n\). If the comparison series diverges, that upper bound does not settle the behavior of the smaller series.

For example, compare the critical model \(a_n=1/[n\log n]\) with \(b_n=1/n\). The ratio is

\frac{a_n}{b_n} = \frac{1}{\log n} \longrightarrow 0.

A mistaken conclusion would be: “The ratio tends to zero, so \(\sum a_n\) converges.” That inference is invalid: the harmonic comparison series \(\sum b_n\) diverges. In fact, the critical series \(\sum 1/[n\log n]\) also diverges, as the integral calculation shows.

Comparison check: A positive finite limit allows the limit comparison test to transfer behavior in either direction. A zero limit gives a convergence conclusion only when the comparison series converges; an infinite limit gives a divergence conclusion only when the comparison series diverges. Other cases require additional reasoning.

A Practical Classification Process

For a series whose terms contain a factor resembling \(1/n\) and a logarithmic power, use the following sequence of decisions.

Identify the logarithmic exponent.
Rewrite roots or fractional powers as \((\log n)^p\), and ensure the terms are positive from some index onward.

Choose the model.
Use \(b_n=1/[n(\log n)^p]\) when it captures the leading behavior of the given term.

Compute the ratio.
Simplify \(a_n/b_n\) and evaluate its limit, showing that it is finite and strictly positive if using ordinary limit comparison.

Apply the threshold.
The model converges for \(p>1\) and diverges for \(p\leq 1\); transfer that result only when the comparison conditions are met.

This method matters because the logarithmic factor can decide the behavior precisely when the power of \(n\) is at its boundary. In the model considered here, the exponent \(1\) is not a harmless endpoint: it belongs to the divergent case. A ratio calculation identifies the right model, while the threshold theorem supplies the classification.

Check Your Understanding

For each question, state the comparison model or threshold reasoning you would use.

For which real values of \(p\) does \(\sum_{n=2}^{\infty}1/[n(\log n)^p]\) converge?
Why does the substitution \(u=\log x\) reduce the integral for the model series to a power integral?
What does a limit of \(a_n/b_n=4\) imply when \(\sum b_n\) converges?
Classify \(\sum_{n=2}^{\infty}(5n+2)/[n^2(\log n)^3]\) by limit comparison. What is the ratio limit with the model?
Why does \(a_n/b_n\to 0\) fail to settle convergence when \(\sum b_n\) diverges?