Iterated Logarithms and Borderline Convergence

Why an Extra Logarithm Changes the Threshold

The harmonic series diverges, and adding a logarithm to its denominator changes the convergence question. With a further logarithmic factor, the answer depends on the exponent of that factor. The transition is especially delicate at one: the series with exponent one still diverges, even though its terms contain two logarithmic factors in the denominator.

The integral test makes this threshold visible. The key is to substitute \(u=\ln\ln x\), because the differential \(du\) contains exactly the factor \(1/(x\ln x)\). After that substitution, the improper integral becomes a power integral in \(u\).

Central conclusion: For real \(p\), the series \(\sum_{n=3}^{\infty} \frac{1}{n\ln n(\ln\ln n)^p}\) converges if \(p>1\) and diverges if \(p\leq 1\).

Where the Terms Are Defined

We consider

\sum_{n=3}^{\infty} \frac{1}{n\ln n\,(\ln\ln n)^p}, \qquad p\in\mathbb{R}.

The starting index matters. For a real power \(a^p\) with arbitrary real \(p\), the base \(a\) must be positive. At every integer \(n\geq 3\), we have \(\ln n>1\), so \(\ln\ln n>0\). Thus the displayed terms are defined and positive for every real \(p\). Starting at \(n=1\) or \(n=2\) would not give the same well-defined expression.

The corresponding function is

f(x)=\frac{1}{x\ln x\,(\ln\ln x)^p}, \qquad x\geq 3.

It is continuous and positive on this interval. To use the integral test, we also need it to be decreasing eventually. For any fixed real \(p\), logarithmic differentiation gives

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

As \(x\) tends to infinity, the quantity in parentheses tends to \(1\). Therefore it is positive for all sufficiently large \(x\), and \(f'(x)<0\) there. This verifies eventual decrease even when \(p\) is negative. The integral test applies to the tail beginning at a sufficiently large integer \(N\); omitting or adding finitely many terms does not change whether a series converges.

Integral Test: If \(f\) is positive, continuous, and decreasing on \([N,\infty)\), then the series \(\sum_{n=N}^{\infty}f(n)\) and the improper integral \(\int_N^\infty f(x)\,dx\) either both converge or both diverge.

The Successive Logarithmic Substitution

Consider the integral from a sufficiently large \(A\geq 3\):

\int_A^\infty \frac{dx}{x\ln x\,(\ln\ln x)^p}.

Set \(u=\ln\ln x\). Differentiating this substitution gives

du=\frac{1}{x\ln x}\,dx.

The lower endpoint becomes \(u=\ln\ln A>0\). If the upper endpoint is a finite number \(R>A\), it becomes \(u=\ln\ln R\). Consequently, the finite-endpoint integral is

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

This equality is the decisive step: both logarithmic factors are accounted for, and the transformed integral has a familiar power integrand. Taking \(R\) to infinity makes \(\ln\ln R\) tend to infinity, so the convergence question is determined by the upper limit of the power integral.

Substitution rule for this problem: With \(u=\ln\ln x\), the factor \(dx/(x\ln x)\) becomes \(du\), and \((\ln\ln x)^{-p}\) becomes \(u^{-p}\). The transformed lower limit is positive when \(A>e\).

Classifying the Improper Integral

There are two cases for evaluating the transformed integral. When \(p\ne 1\), an antiderivative of \(u^{-p}\) is \(u^{1-p}/(1-p)\). When \(p=1\), the antiderivative is \(\ln u\). Thus, for \(R>A\),

\int_A^R \frac{dx}{x\ln x\,(\ln\ln x)^p} = \begin{cases} \displaystyle \frac{(\ln\ln R)^{1-p}-(\ln\ln A)^{1-p}}{1-p}, & p\ne 1,\\[1.2em] \displaystyle \ln(\ln\ln R)-\ln(\ln\ln A), & p=1. \end{cases}

For \(p>1\), the exponent \(1-p\) is negative. Hence \((\ln\ln R)^{1-p}\) tends to zero as \(R\) tends to infinity, and the integral has the finite value

\int_A^\infty \frac{dx}{x\ln x\,(\ln\ln x)^p} = \frac{(\ln\ln A)^{1-p}}{p-1}, \qquad p>1.

For \(p<1\), the exponent \(1-p\) is positive, so \((\ln\ln R)^{1-p}\) tends to infinity. The integral diverges. At the boundary \(p=1\), the expression \(\ln(\ln\ln R)\) also tends to infinity, so the integral diverges there as well.

Integral classification: The improper integral converges exactly when \(p>1\). It diverges for \(p\leq 1\), including the borderline case \(p=1\).

Applying the Classification to the Series

The function \(f\) is positive and continuous for \(x\geq 3\), and the derivative calculation shows that it is decreasing for all sufficiently large \(x\). The integral test therefore transfers the integral classification directly to the series. The first few terms, before eventual decrease begins if necessary, cannot affect convergence.

\sum_{n=3}^{\infty} \frac{1}{n\ln n\,(\ln\ln n)^p} \begin{cases} \text{converges}, & p>1,\\ \text{diverges}, & p\leq 1. \end{cases}

This result is a borderline test: the exponent \(p\) controls a threshold at exactly one. In particular, merely having a factor of \(\ln\ln n\) in the denominator is not enough. Its power must be strictly greater than one for convergence.

Worked Examples

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

Classify \(\sum_{n=3}^{\infty}\frac{1}{n\ln n(\ln\ln n)^2}\). Here \(p=2>1\), so the general criterion predicts convergence. We verify the integral calculation directly.

\int_3^R\frac{dx}{x\ln x(\ln\ln x)^2} = \int_{\ln\ln 3}^{\ln\ln R}u^{-2}\,du = \left[-\frac{1}{u}\right]_{\ln\ln 3}^{\ln\ln R} = \frac{1}{\ln\ln 3}-\frac{1}{\ln\ln R}.

The antiderivative is correct because \(\frac{d}{du}(-u^{-1})=u^{-2}\). As \(R\) tends to infinity, \(\ln\ln R\) tends to infinity, so \(1/\ln\ln R\) tends to zero. The integral therefore converges to \(1/\ln\ln 3\), a finite number. The function is positive and decreasing for \(x\geq3\) when \(p=2\), as the derivative factor is \(1+1/\ln x+2/(\ln x\ln\ln x)>0\). By the integral test, the series converges.

Worked Example: The Borderline Series at \(p=1\)

Classify \(\sum_{n=3}^{\infty}\frac{1}{n\ln n\ln\ln n}\). The transformed integral has integrand \(u^{-1}\):

\int_3^R\frac{dx}{x\ln x\ln\ln x} = \int_{\ln\ln 3}^{\ln\ln R}\frac{du}{u} = \left[\ln u\right]_{\ln\ln 3}^{\ln\ln R} = \ln(\ln\ln R)-\ln(\ln\ln 3).

The antiderivative checks because \(\frac{d}{du}\ln u=1/u\) for \(u>0\). As \(R\) tends to infinity, \(\ln\ln R\) tends to infinity, and then \(\ln(\ln\ln R)\) also tends to infinity. The integral diverges. For this positive exponent the integrand function is decreasing on \(x\geq3\), so the integral test proves that the series diverges. This is the borderline case: the exponent is not large enough for convergence.

Worked Example: Divergence at \(p=\frac12\)

Consider \(\sum_{n=3}^{\infty}\frac{1}{n\ln n\sqrt{\ln\ln n}}\). Here \(p=\frac12<1\). Substitution gives

\int_3^R \frac{dx}{x\ln x\sqrt{\ln\ln x}} = \int_{\ln\ln 3}^{\ln\ln R}u^{-1/2}\,du = \left[2\sqrt{u}\right]_{\ln\ln 3}^{\ln\ln R} = 2\sqrt{\ln\ln R}-2\sqrt{\ln\ln 3}.

The antiderivative checks since \(\frac{d}{du}(2\sqrt{u})=u^{-1/2}\) for \(u>0\). The first term on the right grows without bound as \(R\) tends to infinity, so the integral diverges. The function is positive and decreasing for \(x\geq3\), and the integral test therefore shows that the series diverges.

A Common Antiderivative Error

At \(p=1\), it is easy to stop one logarithm too early. A mistaken calculation might be

\int\frac{dx}{x\ln x\ln\ln x} \ \overset{\text{wrong}}{=}\ \ln\ln x.

Differentiating the proposed answer exposes the error:

\frac{d}{dx}\ln\ln x=\frac{1}{x\ln x},

which is missing the factor \(1/\ln\ln x\) from the integrand. The correction is to substitute \(u=\ln\ln x\), not \(u=\ln x\). Then the integrand becomes \(1/u\), whose antiderivative is \(\ln u\). Returning to \(x\) gives

\ln u=\ln(\ln\ln x), \qquad \frac{d}{dx}\ln(\ln\ln x) = \frac{1}{x\ln x\ln\ln x}.

The nested parentheses matter: the correct antiderivative contains the logarithm of \(\ln\ln x\). Differentiating the proposed antiderivative is a quick and reliable check.

Why the Boundary Case Matters

For \(p>1\), the transformed integrand \(u^{-p}\) has enough decay to produce a finite integral. At \(p=1\), its antiderivative is logarithmic and still grows without bound. For \(p<1\), the power antiderivative grows without bound. This explains why “one more logarithm” does not by itself guarantee convergence: the decisive question is whether the exponent on that logarithmic factor exceeds one.

The classification also covers negative \(p\). In that case, \((\ln\ln n)^p\) appears in the denominator but becomes smaller as \(\ln\ln n\) grows; equivalently, the terms are larger than the corresponding terms with \(p=0\) for sufficiently large \(n\). The integral calculation already includes these values and shows divergence for every \(p<1\).

Check Your Understanding

Use the substitution \(u=\ln\ln x\) and the integral test where appropriate.

For which real values of \(p\) does the series \(\sum_{n=3}^{\infty}\frac{1}{n\ln n(\ln\ln n)^p}\) converge?
Why is \(n=3\) a valid starting index when \(p\) may be any real number?
What does the integral become after setting \(u=\ln\ln x\)?
Why does the case \(p=1\) diverge?
What derivative check identifies the error in claiming that \(\ln\ln x\) is an antiderivative when \(p=1\)?