Integral-Test Workshop: Power and Logarithm Tails

Three Families, Three Thresholds

Power and logarithmic factors can make a series look similar while producing different convergence behavior. The Integral Test turns these problems into improper integrals, but only after its hypotheses have been checked. In particular, borderline exponents require care: a small change in an exponent can change a finite integral into a divergent one.

This tutorial classifies three families: \(1/n^p\), \(1/[n(\ln n)^p]\), and \(1/[n\ln n(\ln\ln n)^p]\). Throughout, \(p\) is a real parameter. For the logarithmic families, the starting indices are chosen so that every logarithm appearing as a base is positive.

Integral Test: Suppose \(f\) is positive, continuous, and decreasing on \([N,\infty)\), where \(N\) is an integer. Then the series \(\sum_{n=N}^{\infty} f(n)\) converges if and only if the improper integral \(\int_N^\infty f(x)\,dx\) is finite. If the integral is infinite, the series diverges. Adding or removing finitely many terms does not change whether a series converges.

Why the Integral Test Works

The reason for the test is the comparison between the areas under a decreasing curve and rectangles of width \(1\). If \(f\) is decreasing, then on \([n,n+1]\), its values are at most \(f(n)\). On \([n-1,n]\), its values are at least \(f(n)\). Therefore, for integers \(M\geq N\),

\int_{N+1}^{M+1} f(x)\,dx \leq \sum_{n=N}^{M} f(n) \leq f(N)+\int_N^M f(x)\,dx.

The left inequality follows by applying \(f(x)\leq f(n)\) on each interval \([n,n+1]\), for \(n=N,\ldots,M\). For the right inequality, separate the first term \(f(N)\); on each interval \([n-1,n]\), \(f(x)\geq f(n)\), so \(f(n)\leq\int_{n-1}^{n} f(x)\,dx\) for \(n=N+1,\ldots,M\). These bounds show that the partial sums stay bounded exactly when the integrals over the corresponding tails stay bounded. Since the terms are positive, bounded partial sums are equivalent to convergence. This proves the test.

In applications, a function may be decreasing only after some point. That is enough: discard the finite initial segment and apply the test to the remaining tail. The discarded terms have a finite sum and cannot change convergence or divergence.

The Power Family: \(1/n^p\)

For \(p>0\), the function \(f(x)=x^{-p}\) is positive and continuous for \(x\geq1\), and its derivative is \(-p x^{-p-1}<0\). It is therefore decreasing, so the Integral Test applies. For \(p\leq0\), the terms \(1/n^p=n^{-p}\) do not tend to zero: they equal \(1\) when \(p=0\), and grow when \(p<0\). Those series diverge.

For \(p\neq1\), the antiderivative is \(x^{1-p}/(1-p)\); when \(p=1\), it is \(\ln x\). If \(p>1\), \(x^{1-p}\) tends to zero as \(x\) tends to infinity, so the improper integral is finite. If \(0<p<1\), \(x^{1-p}\) grows without bound, and if \(p=1\), \(\ln x\) grows without bound. Thus the full classification is:

Power-series classification: The series \(\sum_{n=1}^{\infty}1/n^p\) converges exactly when \(p>1\), and diverges when \(p\leq1\).

Worked Example: A Convergent Power Series

Classify \(\sum_{n=1}^{\infty}1/n^{3/2}\). Set \(f(x)=x^{-3/2}\). It is positive and continuous for \(x\geq1\), and

f'(x)=-\frac{3}{2}x^{-5/2}<0.

Thus \(f\) is decreasing and the Integral Test applies. Evaluate the integral:

\int_1^\infty x^{-3/2}\,dx = \lim_{b\to\infty}\left[-2x^{-1/2}\right]_1^b = \lim_{b\to\infty}\left(2-\frac{2}{\sqrt b}\right) = 2.

The integral is finite, so the series converges. As a check on the antiderivative, \(\frac{d}{dx}(-2x^{-1/2})=x^{-3/2}\).

One Logarithmic Factor: \(1/[n(\ln n)^p]\)

For real \(p\), consider \(f(x)=1/[x(\ln x)^p]\) on \(x\geq2\). Here \(\ln x>0\), so this is defined and positive. Its derivative can be written as

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

If \(p\geq0\), the expression in parentheses is positive for \(x\geq2\). If \(p<0\), it is positive whenever \(\ln x>-p\), which holds for all sufficiently large \(x\). Therefore \(f\) is eventually decreasing for every fixed real \(p\). The Integral Test applies to a suitable tail, and any omitted initial terms are finite.

Use \(u=\ln x\), so \(du=dx/x\). The resulting integral has the same form as a power integral:

\int \frac{dx}{x(\ln x)^p} = \int u^{-p}\,du = \begin{cases} \dfrac{(\ln x)^{1-p}}{1-p}+C, & p\neq1,\\[6pt] \ln(\ln x)+C, & p=1. \end{cases}

When \(p>1\), \((\ln x)^{1-p}\) tends to zero, so the integral converges. When \(p<1\), it grows without bound; when \(p=1\), \(\ln(\ln x)\) grows without bound. This includes negative \(p\), for which the antiderivative also tends to infinity.

Logarithmic-series classification: For the tail \(\sum_{n=2}^{\infty}1/[n(\ln n)^p]\), convergence occurs exactly when \(p>1\). The series diverges for \(p\leq1\).

Worked Example: The Logarithmic Borderline

Classify \(\sum_{n=2}^{\infty}1/[n\ln n]\), which corresponds to \(p=1\). The function \(f(x)=1/(x\ln x)\) is positive and continuous for \(x\geq2\). Differentiating gives

f'(x)=-\frac{\ln x+1}{x^2(\ln x)^2}<0 \qquad (x\geq2).

It is decreasing, so the Integral Test applies. Its integral is

\int_2^b \frac{dx}{x\ln x} = \left[\ln(\ln x)\right]_2^b = \ln(\ln b)-\ln(\ln2).

As \(b\to\infty\), \(\ln(\ln b)\to\infty\). The integral diverges, so the series diverges. Differentiating the antiderivative verifies the calculation: \(\frac{d}{dx}\ln(\ln x)=1/(x\ln x)\).

Two Logarithmic Factors: An Iterated-Logarithm Threshold

Now consider \(f(x)=1/[x\ln x(\ln\ln x)^p]\). For \(x\geq3\), both \(\ln x\) and \(\ln\ln x\) are positive, so \(f\) is defined and positive. Its logarithmic derivative is

\frac{f'(x)}{f(x)} = -\frac{1}{x} \left( 1+\frac{1}{\ln x} +\frac{p}{\ln x\,\ln\ln x} \right).

For any fixed real \(p\), the expression in parentheses tends to \(1\) as \(x\to\infty\). It is therefore positive for all sufficiently large \(x\), which makes \(f\) eventually decreasing. This is enough to apply the Integral Test to the tail of the series starting at \(n=3\).

Set \(u=\ln\ln x\). Then \(du=dx/(x\ln x)\), and the integral becomes \(\int u^{-p}\,du\). Its evaluation is

\int \frac{dx}{x\ln x(\ln\ln x)^p} = \begin{cases} \dfrac{(\ln\ln x)^{1-p}}{1-p}+C, & p\neq1,\\[6pt] \ln(\ln\ln x)+C, & p=1. \end{cases}

Since \(\ln\ln x\to\infty\), this integral is finite exactly when \(p>1\). For \(p<1\), the power \((\ln\ln x)^{1-p}\) grows without bound; for \(p=1\), \(\ln(\ln\ln x)\) grows without bound.

Iterated-logarithm classification: The series \(\sum_{n=3}^{\infty}1/[n\ln n(\ln\ln n)^p]\) converges exactly when \(p>1\), and diverges when \(p\leq1\).

Worked Example: A Convergent Iterated-Logarithm Series

Classify \(\sum_{n=3}^{\infty}1/[n\ln n(\ln\ln n)^2]\). Let \(f(x)=1/[x\ln x(\ln\ln x)^2]\). It is positive and continuous for \(x\geq3\). For \(p=2\), the derivative calculation above gives

\frac{f'(x)}{f(x)} = -\frac{1}{x} \left( 1+\frac{1}{\ln x} +\frac{2}{\ln x\,\ln\ln x} \right)<0 \qquad (x\geq3).

Thus \(f\) is decreasing. Using \(u=\ln\ln x\), evaluate the integral from \(3\) to infinity:

\int_3^\infty \frac{dx}{x\ln x(\ln\ln x)^2} = \lim_{b\to\infty} \left[-\frac{1}{\ln\ln x}\right]_3^b = \frac{1}{\ln\ln3}.

Because \(\ln\ln3>0\), this is a finite positive number. The series converges by the Integral Test. The antiderivative checks because \(\frac{d}{dx}[-1/\ln\ln x]=1/[x\ln x(\ln\ln x)^2]\).

Compare the Thresholds

In each family, the threshold is \(p=1\), but the factors preceding the exponent differ. At the threshold, each new logarithmic factor produces another borderline integral. The integral calculation, rather than the visual appearance of the terms, determines the classification.

Series tail	Converges when	Diverges when
\(\sum 1/n^p\)	\(p>1\)	\(p\leq1\)
\(\sum 1/[n(\ln n)^p]\)	\(p>1\)	\(p\leq1\)
\(\sum 1/[n\ln n(\ln\ln n)^p]\)	\(p>1\)	\(p\leq1\)

A Common Mistake: Skipping the Hypothesis Check

A tempting but incomplete argument is: “The integral has a finite antiderivative, so the series converges.” An antiderivative alone says nothing about whether the improper integral is finite, and the Integral Test also requires a positive, continuous, decreasing function on the tail.

For example, when \(p=1\), one might compute \(\int dx/(x\ln x)=\ln(\ln x)\) and stop. The correction is to evaluate the improper limit:

\int_2^\infty \frac{dx}{x\ln x} = \lim_{b\to\infty} \left(\ln(\ln b)-\ln(\ln2)\right) = \infty.

The integral is divergent, and only after checking that \(1/(x\ln x)\) is positive, continuous, and decreasing on \([2,\infty)\) can the Integral Test be used to conclude that the series diverges.

How to Organize an Integral-Test Classification

For these families, a reliable solution has three parts: verify the test's conditions on the chosen tail, evaluate the improper integral including its limit at infinity, and translate the integral's behavior into the series classification. The substitutions \(u=\ln x\) and \(u=\ln\ln x\) reveal the relevant power integral; the endpoint limit decides whether it is finite.

Choose a valid tail.
Make sure the function is defined and positive there; for iterated logarithms, \(x\geq3\) ensures \(\ln\ln x>0\).

Check eventual decrease.
Use the derivative, and if necessary move the starting point farther out. A finite number of omitted terms does not affect the result.

Evaluate the improper integral.
Use the relevant logarithmic substitution, then take the limit at infinity rather than stopping at an antiderivative.

State the conclusion and threshold.
A finite integral gives a convergent series; an infinite integral gives a divergent series.

Check Your Understanding

For each question, identify the relevant integral and state the convergence conclusion with its condition.

For which real values of \(p\) does \(\sum_{n=1}^{\infty}1/n^p\) converge?
What substitution evaluates \(\int dx/[x(\ln x)^p]\), and what happens at \(p=1\)?
Why is eventual decrease sufficient when applying the Integral Test?
Does \(\sum_{n=2}^{\infty}1/[n(\ln n)^{1/2}]\) converge or diverge? Justify using the improper integral.
For which real values of \(p\) does \(\sum_{n=3}^{\infty}1/[n\ln n(\ln\ln n)^p]\) converge?