Integral Tests for Logarithmic Boundary Series

Why Logarithmic Factors Need Their Own Test

A boundary series can have positive terms that look almost like the harmonic series, but with an additional logarithmic factor. That factor can change the convergence verdict: for example, raising the logarithm to a power greater than one gives a different outcome from raising it to the first power.

The integral test gives a direct way to identify the threshold. Its conclusion is reliable only after its hypotheses have been checked. In this case, the main details are that the terms are positive and continuous when viewed as a function, and that the function is decreasing at least eventually.

Scope: We study the positive series whose terms are \(1/[n(\log n)^q]\), beginning at \(n=2\), where \(q\) is any real number and \(\log\) denotes the natural logarithm. Beginning at \(n=2\) keeps the logarithm positive and avoids the undefined term at \(n=1\).

The Integral Test and Its Conditions

The integral test relates a series of positive terms to the area under a function. It requires more than a formula that resembles an integral: the function must be positive, continuous, and decreasing on the relevant interval.

Theorem (Integral Test): Suppose \(f\) is positive, continuous, and decreasing on \([N,\infty)\), where \(N\) is a positive integer, and suppose \(a_n=f(n)\) for integers \(n\ge N\). Then \(\sum_{n=N}^{\infty}a_n\) converges if and only if the improper integral \(\int_N^\infty f(x)\,dx\) converges. The same conclusion holds if these conditions are satisfied only from some integer onward, because changing finitely many terms does not change convergence.

For the series in this tutorial, associate the function

f(x)=\frac{1}{x(\log x)^q},\qquad x>1.

For every real \(q\), \(f(x)\) is positive on \(x>1\): the denominator \(x(\log x)^q\) is positive there. It is also continuous on \(x>1\), since \(x\), \(\log x\), and the real power of the positive quantity \(\log x\) are continuous on that domain. We still need to establish where \(f\) is decreasing.

Logarithmic differentiation, or the product and chain rules, gives

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

The denominator in this expression is positive for \(x>1\), even when \(q+1\) is negative. Thus \(f'(x)<0\) whenever \(\log x+q>0\), which is equivalent to \(x>e^{-q}\). In particular, for any real \(q\), we can choose an integer \(N>\max\{1,e^{-q}\}\); then \(f\) is positive, continuous, and decreasing on \([N,\infty)\). This eventual decrease is enough for the integral test. The first few terms, before \(N\), have no effect on whether the infinite series converges.

Why the Integral Test Works

Here is the comparison behind the theorem. If \(f\) is decreasing, then on each interval \([k,k+1]\), its values are at most \(f(k)\) and at least \(f(k+1)\). Therefore

f(k+1)\le \int_k^{k+1}f(x)\,dx\le f(k).

Adding these inequalities over consecutive intervals gives bounds between partial sums and integrals. In particular, for integers \(M>N\),

\int_N^{M+1}f(x)\,dx \le \sum_{k=N}^{M}f(k)

and

\sum_{k=N+1}^{M}f(k) \le \int_N^M f(x)\,dx.

If the improper integral is finite, the second inequality bounds the partial sums of the series, so the positive series converges. If the improper integral is infinite, the first inequality shows that the partial sums of the series are unbounded, so the series diverges. This proves the integral-test conclusion for positive decreasing functions. The inequalities also explain why the direction of each comparison matters.

Evaluating the Logarithmic Integral

For the logarithmic function in question, the substitution \(u=\log x\) gives \(du=dx/x\). Thus the improper integral is governed by

\int \frac{dx}{x(\log x)^q} =\int u^{-q}\,du.

There are two cases because the power rule for integration has a separate logarithmic case when the exponent of \(u\) is \(-1\). If \(q\ne1\), an antiderivative is \((\log x)^{1-q}/(1-q)\). If \(q=1\), an antiderivative is \(\log(\log x)\). Applying the upper limit at infinity yields the convergence threshold.

Conclusion: For every real \(q\), the positive series \(\sum_{n=2}^{\infty}1/[n(\log n)^q]\) converges exactly when \(q>1\). It diverges when \(q\le1\). The integral test is applicable because the associated function is positive and continuous for \(x>1\), and decreases for all sufficiently large \(x\).

For \(q>1\), the exponent \(1-q\) is negative, so \((\log x)^{1-q}\) tends to zero as \(x\) tends to infinity. The improper integral is finite. For \(q<1\), the exponent \(1-q\) is positive, so that power tends to infinity. At \(q=1\), \(\log(\log x)\) also tends to infinity. In both cases \(q\le1\), the improper integral diverges.

Worked Examples

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

Determine whether \(\sum_{n=2}^{\infty}1/[n(\log n)^2]\) converges.

Use \(f(x)=1/[x(\log x)^2]\). For \(x>1\), this function is positive and continuous. Its derivative is

f'(x)=-\frac{\log x+2}{x^2(\log x)^3}<0,

because every factor in the denominator is positive and \(\log x+2>0\) for \(x>1\). Thus the integral test applies from \(x=2\). Evaluate the integral:

\int_2^\infty\frac{dx}{x(\log x)^2} =\lim_{b\to\infty}\left[-\frac{1}{\log x}\right]_{2}^{b} =\lim_{b\to\infty}\left(\frac{1}{\log 2}-\frac{1}{\log b}\right) =\frac{1}{\log 2}<\infty.

To check the antiderivative, differentiate \(-1/\log x\): its derivative is \(1/[x(\log x)^2]\). The integral is finite, so the series converges by the integral test.

Worked Example: The Threshold Case \(q=1\)

Determine whether \(\sum_{n=2}^{\infty}1/[n\log n]\) converges.

Set \(f(x)=1/[x\log x]\). It is positive and continuous for \(x>1\), and

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

Therefore the integral test applies. An antiderivative is \(\log(\log x)\), since

\frac{d}{dx}\log(\log x)=\frac{1}{x\log x}.

The improper integral is

\int_2^\infty\frac{dx}{x\log x} =\lim_{b\to\infty}\bigl(\log(\log b)-\log(\log 2)\bigr) =\infty.

Hence the series diverges. This is the threshold case: the logarithmic factor has power one, not a power greater than one.

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

Determine whether \(\sum_{n=2}^{\infty}1/[n\sqrt{\log n}]\) converges.

Here \(f(x)=1/[x(\log x)^{1/2}]\), which is positive and continuous for \(x>1\). Its derivative has the sign given by

f'(x)=-\frac{\log x+\tfrac12}{x^2(\log x)^{3/2}}<0 \qquad (x>1).

So the integral test applies on \([2,\infty)\). Using \(u=\log x\), or differentiating the resulting antiderivative directly, gives

\int_2^b\frac{dx}{x(\log x)^{1/2}} =2\sqrt{\log b}-2\sqrt{\log 2}.

Indeed, the derivative of \(2\sqrt{\log x}\) is \(1/[x\sqrt{\log x}]\). As \(b\to\infty\), the integral tends to infinity. The series therefore diverges, as predicted for \(q<1\).

Worked Example: A Negative Exponent and Eventual Decrease

Determine whether \(\sum_{n=2}^{\infty}(\log n)^2/n\) converges. This is the case \(q=-2\), since \(1/[n(\log n)^{-2}]=(\log n)^2/n\).

The associated function \(f(x)=(\log x)^2/x\) is positive and continuous for \(x>1\). Its derivative is

f'(x)=\frac{\log x(2-\log x)}{x^2}.

It is negative when \(\log x>2\), or \(x>e^2\). Thus \(f\) is decreasing from any integer \(N>e^2\) onward; it need not be decreasing on the whole interval starting at \(2\). This eventual decrease is sufficient. An antiderivative is \((\log x)^3/3\), because

\frac{d}{dx}\left(\frac{(\log x)^3}{3}\right) =\frac{(\log x)^2}{x}.

The integral from \(N\) to \(b\) is \(((\log b)^3-(\log N)^3)/3\), which tends to infinity. The integral test therefore shows that the series diverges.

Using the Result at a Series Boundary

When a power series is evaluated at an endpoint and its terms become positive expressions of the form \(1/[n(\log n)^q]\), this test gives the endpoint verdict: the resulting series converges for \(q>1\) and diverges for \(q\le1\). The integral calculation decides the outcome only after the endpoint substitution has produced the stated positive terms. If the terms have signs that alternate, or are otherwise not positive, the integral test cannot be applied directly to that signed series.

The test also permits omitting a finite number of terms. For instance, the function may fail to decrease for small \(x\) when \(q\) is negative, but it eventually decreases. Removing or adding the finitely many terms before that point does not affect convergence. What matters is that the hypotheses hold on some entire interval \([N,\infty)\), not necessarily from the first index of the written series.

A Common Mistake

A frequent error is to treat the threshold case \(q=1\) as though the logarithm made the harmonic-type terms small enough to converge. The integral shows why that conclusion is false:

\int_2^\infty\frac{dx}{x\log x} =\lim_{b\to\infty}\bigl(\log(\log b)-\log(\log 2)\bigr) =\infty.

So the integral test gives divergence, not convergence. Another possible mistake is to apply the test without checking monotonicity. For a particular \(q\), the function need not decrease from \(x=2\), but the derivative formula shows it decreases once \(\log x+q>0\). State the interval on which the hypotheses hold, then apply the test to that tail of the series.

Key takeaway: The logarithmic power must be strictly greater than one for convergence. Positivity and continuity hold for \(x>1\), while decrease is guaranteed eventually for every real exponent \(q\).

Check Your Understanding

For each question, justify your answer using the integral test and its conditions where appropriate.

For which real values of \(q\) does \(\sum_{n=2}^{\infty}1/[n(\log n)^q]\) converge?
What derivative condition shows that \(1/[x(\log x)^q]\) is eventually decreasing?
Why is it enough for the associated function to decrease only after some integer \(N\)?
Evaluate the improper integral that corresponds to \(q=1\), and state its convergence verdict.
Can the integral test be applied directly to a series whose terms alternate in sign? Explain why or why not.