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Integral and Direct Comparison · Tutorial 604 of 998

Logarithmic Factors in Harmonic-Type Series

Learn how logarithmic factors change the convergence threshold for a harmonic-type series and how to justify the classification for every real exponent.

Intermediate 9 min read

What You'll Learn

  • State when the integral test applies to logarithmic terms.
  • Check eventual decrease for every real exponent.
  • Use the substitution u = ln x to evaluate the relevant improper integral.
  • Explain why the convergence threshold is p = 1.
  • Recognize and correct a common substitution error.

Why a Logarithm Changes the Harmonic Series

The harmonic series \(\sum_{n=1}^{\infty}1/n\) diverges. A natural question is whether dividing its terms by a power of a logarithm can make the sum converge. The answer depends precisely on the exponent: for the series considered here, the boundary between convergence and divergence occurs at \(p=1\).

We will establish this classification with the integral test. The key step is the substitution \(u=\ln x\), which turns the factor \(dx/x\) into \(du\) and converts the logarithmic factor into a power of \(u\).

Definition: For a real number \(p\), consider the series $$ \sum_{n=2}^{\infty}\frac{1}{n(\ln n)^p}. $$ For a positive base \(a\) and real exponent \(p\), \(a^p\) has its usual real-valued meaning. Here \(\ln n>0\) for every index \(n\ge 2\), so each term is defined and positive for every real \(p\).

Using the Integral Test

The integral test connects a positive, decreasing sequence of terms to the area under a corresponding function. In the form we need, if \(f\) is continuous, positive, and decreasing on \([N,\infty)\), and \(a_n=f(n)\) for integers \(n\ge N\), then \(\sum_{n=N}^{\infty}a_n\) converges exactly when the improper integral \(\int_N^\infty f(x)\,dx\) converges. A finite number of initial terms does not affect whether a series converges.

For this problem, take

$$ f(x)=\frac{1}{x(\ln x)^p}, \qquad x\ge 2. $$

The function is continuous and positive on this domain. Its decrease needs a little care because \(p\) may be negative. Differentiating gives

$$ \begin{aligned} f'(x) &=-\frac{1}{x^2(\ln x)^p} -\frac{p}{x^2(\ln x)^{p+1}}\\ &=-\frac{1+\dfrac{p}{\ln x}}{x^2(\ln x)^p}. \end{aligned} $$

The denominator is positive. If \(p\ge 0\), the numerator \(1+p/\ln x\) is positive for \(x\ge 2\). If \(p<0\), it is positive whenever \(\ln x>-p\), which holds for all sufficiently large \(x\). Thus \(f\) is eventually decreasing for every real \(p\). The integral test therefore applies to a tail of the series, regardless of the value of \(p\).

The Substitution That Reveals the Threshold

Consider the improper integral starting at any \(A>1\) large enough that \(f\) is decreasing. For a finite upper endpoint \(B>A\), set \(u=\ln x\). Then \(du=dx/x\), and the endpoints become \(u=\ln A\) and \(u=\ln B\). Consequently,

$$ \int_A^B\frac{dx}{x(\ln x)^p} = \int_{\ln A}^{\ln B}u^{-p}\,du. $$

This equality follows directly from \(du=dx/x\) and \((\ln x)^{-p}=u^{-p}\). Letting \(B\) tend to infinity makes the upper endpoint \(\ln B\) tend to infinity as well. The question is therefore whether \(\int_{\ln A}^{\infty}u^{-p}\,du\) converges.

Result: The series \(\sum_{n=2}^{\infty}1/[n(\ln n)^p]\) converges if and only if \(p>1\). It diverges for every \(p\le 1\).

Why the Cutoff Is Exactly \(p=1\)

We can determine the integral’s behavior by evaluating it for the two cases \(p\ne 1\) and \(p=1\). If \(p\ne 1\), an antiderivative of \(u^{-p}\) is \(u^{1-p}/(1-p)\), since differentiating this expression gives \(u^{-p}\). Thus, for \(T>\ln A>0\),

$$ \int_{\ln A}^{T}u^{-p}\,du = \frac{T^{1-p}-(\ln A)^{1-p}}{1-p}. $$

If \(p>1\), then \(1-p<0\), so \(T^{1-p}\) tends to zero as \(T\) tends to infinity. The improper integral is finite, with value \((\ln A)^{1-p}/(p-1)\). If \(p<1\), then \(1-p>0\), so \(T^{1-p}\) grows without bound and the integral diverges.

At \(p=1\), the integrand is \(1/u\), whose antiderivative is \(\ln u\) for \(u>0\). Hence

$$ \int_{\ln A}^{T}\frac{du}{u} = \ln T-\ln(\ln A), $$

which tends to infinity as \(T\) tends to infinity. This gives the complete integral classification. Since the integral test applies to the series tail, it gives the same classification for the series.

1
Choose the function.
Use \(f(x)=1/[x(\ln x)^p]\), which is positive and continuous for \(x\ge 2\).
2
Check decrease.
The derivative shows that \(f\) is decreasing for all sufficiently large \(x\), even when \(p\) is negative.
3
Substitute \(u=\ln x\).
The integral becomes \(\int u^{-p}\,du\), with an upper endpoint tending to infinity.
4
Classify the power integral.
It converges for \(p>1\) and diverges for \(p\le 1\); the integral test transfers that result to the series.

Worked Examples

Worked Example: A Convergent Series with \(p=2\)

Classify \(\sum_{n=2}^{\infty}1/[n(\ln n)^2]\). Here \(p=2\), and the associated function is positive, continuous, and decreasing for \(x\ge 2\), as also follows from \(1+2/\ln x>0\). Evaluate the integral from \(2\) to \(B>2\):

$$ \begin{aligned} \int_2^B\frac{dx}{x(\ln x)^2} &=\int_{\ln 2}^{\ln B}u^{-2}\,du\\ &=\left[-u^{-1}\right]_{\ln 2}^{\ln B}\\ &=-\frac{1}{\ln B}+\frac{1}{\ln 2}. \end{aligned} $$

As \(B\to\infty\), \(1/\ln B\to 0\), so the improper integral equals \(1/\ln 2\), a finite number. By the integral test, the series converges. Check the antiderivative in the original variable: \(\frac{d}{dx}[-1/\ln x]=1/[x(\ln x)^2]\), as required.

Worked Example: The Borderline Case \(p=1\)

Classify \(\sum_{n=2}^{\infty}1/(n\ln n)\). The function \(1/(x\ln x)\) is positive and continuous for \(x\ge2\), and its derivative is \(-(\ln x+1)/[x^2(\ln x)^2]<0\). For \(B>2\), the substitution gives

$$ \begin{aligned} \int_2^B\frac{dx}{x\ln x} &=\int_{\ln 2}^{\ln B}\frac{du}{u}\\ &=\left[\ln u\right]_{\ln 2}^{\ln B}\\ &=\ln(\ln B)-\ln(\ln 2). \end{aligned} $$

This expression tends to infinity as \(B\to\infty\), so the improper integral diverges and the integral test shows that the series diverges. The antiderivative checks directly: for \(x>1\), \(\frac{d}{dx}\ln(\ln x)=1/(x\ln x)\).

Worked Example: A Divergent Series with \(p=\tfrac12\)

Classify \(\sum_{n=2}^{\infty}1/[n\sqrt{\ln n}]\). The associated function is positive and continuous for \(x\ge2\); since \(p=1/2\ge0\), its derivative is negative there. For \(B>2\),

$$ \begin{aligned} \int_2^B\frac{dx}{x(\ln x)^{1/2}} &=\int_{\ln 2}^{\ln B}u^{-1/2}\,du\\ &=\left[2u^{1/2}\right]_{\ln 2}^{\ln B}\\ &=2\sqrt{\ln B}-2\sqrt{\ln 2}. \end{aligned} $$

Because \(\ln B\to\infty\), this integral tends to infinity. The integral test therefore shows that the series diverges. Differentiating \(2\sqrt{\ln x}\) gives \(2\cdot\frac12(\ln x)^{-1/2}\cdot1/x=1/[x\sqrt{\ln x}]\), verifying the antiderivative.

Worked Example: A Negative Exponent, \(p=-1\)

Classify \(\sum_{n=2}^{\infty}(\ln n)/n\). This is the case \(p=-1\), because \((\ln n)^{-p}=\ln n\). The function \(f(x)=(\ln x)/x\) has derivative \((1-\ln x)/x^2\), which is negative for \(x>e\). It is therefore eventually decreasing, so the integral test applies to its tail. For \(B>e^2\),

$$ \begin{aligned} \int_{e^2}^B\frac{\ln x}{x}\,dx &=\int_2^{\ln B}u\,du\\ &=\left[\frac{u^2}{2}\right]_2^{\ln B}\\ &=\frac{(\ln B)^2}{2}-2. \end{aligned} $$

This tends to infinity, so the series diverges. As a check, differentiating \(\tfrac12(\ln x)^2\) gives \((\ln x)/x\). This example also shows why the decrease condition must be checked eventually rather than assumed to hold from the first index.

A Substitution Pitfall

The substitution \(u=\ln x\) gives \(du=dx/x\), not \(du/u=dx/x\). Confusing these differentials changes the integral and can give the wrong convergence result. For example, applying the incorrect replacement \(dx/x=du/u\) to the general integrand would lead to the false expression

$$ \frac{dx}{x(\ln x)^p} \overset{\text{incorrect}}{=} \frac{du}{u^{p+1}}. $$

The correction is to replace \(dx/x\) by \(du\) and \(\ln x\) by \(u\):

$$ \frac{dx}{x(\ln x)^p} = u^{-p}\,du. $$

In the borderline case \(p=1\), the correct transformed integral is \(\int du/u\), which diverges logarithmically. In particular, \(\int dx/(x\ln x)=\int du/u=\ln u+C=\ln(\ln x)+C\) for \(x>1\). Differentiating the final expression confirms that it is the correct antiderivative.

Why the Threshold Matters

A logarithmic factor changes the rate of decay of the harmonic term \(1/n\), but it does not behave like an extra fixed power of \(n\). The integral test makes the distinction precise: after substitution, the issue becomes the familiar convergence threshold for the power integral \(\int u^{-p}\,du\). At \(p=1\), the integral becomes \(\int du/u\), and that borderline case still diverges.

Keep the conclusion tied to the exponent in the logarithm: \(p>1\) gives convergence, while \(p=1\) and every smaller real value give divergence. Negative \(p\) is allowed; it makes the logarithm a factor in the numerator and does not invalidate the test, provided eventual decrease is checked.

Check Your Understanding

Use the integral-test reasoning in this tutorial to answer the following questions.

  1. For which real values of \(p\) does \(\sum_{n=2}^{\infty}1/[n(\ln n)^p]\) converge?
  2. With \(u=\ln x\), what differential replaces \(dx/x\), and what does the integrand become?
  3. Why is eventual decrease enough to use the integral test when \(p<0\)?
  4. Evaluate \(\int_2^B dx/[x(\ln x)^3]\) and determine its limit as \(B\to\infty\).
  5. Explain why the case \(p=1\) diverges even though the terms include a logarithmic factor in the denominator.

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