Why logarithmic series need careful checks
Logarithmic factors can change the convergence of a series even when its terms have a similar overall shape. For example, adding a power of \(\ln n\) to a denominator may turn a divergent series into a convergent one. The integral test can resolve these cases, but only after its hypotheses are checked.
For logarithmic terms, the main diagnostic work is often to identify where the corresponding function is positive and continuous, and whether it decreases eventually. Once those conditions hold, substitutions such as \(u=\ln x\) or \(u=\ln\ln x\) turn the improper integral into a familiar power integral.
Diagnosing the hypotheses
A logarithm imposes domain restrictions. For example, \(\ln x\) is positive only when \(x>1\), while \(\ln\ln x\) is positive when \(x>e\). A real power such as \((\ln x)^p\), for an arbitrary real \(p\), is straightforward to use as a positive real-valued factor when \(\ln x>0\). Before applying the integral test, choose a starting point beyond all relevant restrictions.
The test requires decrease only on a tail, not necessarily from the first term of the series. A function might increase over part of its domain and still decrease for every sufficiently large \(x\). In that case, the integral test applies to the tail, and the finite number of earlier terms do not affect the conclusion.
For a function built from several positive factors, monotonicity may follow because its denominator increases. When that is not apparent, differentiating is a reliable check. In logarithmic expressions, the sign of the derivative—not simply the fact that a logarithm appears—determines whether the function decreases.
A convergence threshold from one logarithm
Consider the family of series with terms \(a_n=1/[n(\ln n)^p]\), beginning at \(n=2\), where \(p\) is a real parameter. The associated function is positive and continuous for \(x>1\). Its derivative is
For any fixed real \(p\), the factor \(1+p/\ln x\) is positive for all sufficiently large \(x\), since it approaches \(1\) as \(x\) increases. Thus \(f\) is eventually decreasing, including when it may not decrease near \(x=1\). The integral test therefore applies to a suitable tail.
To evaluate the integral, set \(u=\ln x\), so \(du=dx/x\). The lower endpoint \(x=2\) becomes \(u=\ln 2\), and the upper endpoint tends to infinity:
If \(p\ne 1\), an antiderivative is \(u^{1-p}/(1-p)\). If \(p=1\), an antiderivative is \(\ln u\). Consequently, the integral is finite exactly when \(p>1\). This gives a convergence threshold for the whole family.
Worked Example: A convergent series with a logarithmic power
Worked Example: Testing \(\sum_{n=2}^{\infty}1/[n(\ln n)^{3/2}]\)
Let \(f(x)=1/[x(\ln x)^{3/2}]\). For \(x\ge2\), \(x>0\) and \(\ln x>0\), so \(f\) is positive and continuous. Using the derivative formula above with \(p=3/2\),
Thus the integral test applies. Substitute \(u=\ln x\), with \(du=dx/x\):
The integral is finite, so the series converges. As a check on the antiderivative, differentiating \(-2(\ln x)^{-1/2}\) gives \(1/[x(\ln x)^{3/2}]\), the original integrand.
Iterating the logarithm
A second logarithm can produce another threshold. For \(x>e\), both \(\ln x\) and \(\ln\ln x\) are positive. Consider
For every fixed real \(q\), this function is positive and continuous on a sufficiently far-right interval. To check decrease, differentiate its logarithm:
The right-hand side is negative for all sufficiently large \(x\): after factoring out \(-1/x\), the remaining factor approaches \(1\). Because \(g(x)>0\), \(g'(x)<0\) there. This verifies eventual decrease even for negative values of \(q\), for which the denominator-factor shortcut would not apply.
For the integral, substitute \(u=\ln\ln x\). Since \(du=dx/(x\ln x)\), the resulting integral is \(\int u^{-q}\,du\), with a positive lower endpoint once \(x>e\). It converges exactly when \(q>1\). The borderline case \(q=1\) gives the divergent integral \(\int du/u\).
Worked Example: A convergent iterated-logarithmic series
Worked Example: Testing \(\sum_{n=3}^{\infty}1/[n\ln n(\ln\ln n)^2]\)
Use \(g(x)=1/[x\ln x(\ln\ln x)^2]\). For \(x\ge3>e\), all three denominator factors are positive and increase with \(x\), so \(g\) is positive, continuous, and decreasing. The integral test applies.
Let \(u=\ln\ln x\), so \(du=dx/(x\ln x)\). The lower endpoint \(x=3\) gives \(u=\ln\ln3>0\):
This is finite, so the series converges. Differentiating \(-1/\ln\ln x\) gives \(1/[x\ln x(\ln\ln x)^2]\), confirming the antiderivative and the evaluation.
Worked Example: Eventual decrease with a divergent integral
Worked Example: Testing \(\sum_{n=2}^{\infty}(\ln n)/n\)
Set \(f(x)=\ln x/x\). This function is positive and continuous for \(x>1\). Its derivative is
Thus \(f'(x)<0\) when \(x>e\), so \(f\) is decreasing on a sufficiently far-right interval. It need not be decreasing from the very first index: the integral test only requires eventual decrease. Apply it to a tail starting at \(n=3\).
An antiderivative of \(\ln x/x\) is \(\tfrac12(\ln x)^2\). Therefore,
The integral diverges, so the series diverges. Differentiating \(\tfrac12(\ln x)^2\) gives \(\ln x/x\), which checks the antiderivative.
A common logarithmic-integral mistake
A frequent error is to integrate \(1/(x\ln x)\) as though its antiderivative were \(\ln(x\ln x)\). That candidate is incorrect:
which contains an extra \(1/x\) term. The correct substitution is \(u=\ln x\), giving \(du=dx/x\) and therefore
For \(x>1\), \(\ln x>0\), so this can be written \(\ln(\ln x)+C\). Its growth without bound as \(x\to\infty\) explains why the corresponding integral, and the series \(\sum 1/(n\ln n)\), diverges.
Putting the diagnostic together
For a logarithmic series, organize the reasoning in the same order each time. First identify a function whose integer values are the terms. Then find a tail on which all logarithmic factors are defined and positive. Check that the function is continuous and decreasing there, using a derivative if needed. Only then evaluate the improper integral, keeping its upper limit as a limit rather than substituting infinity into an antiderivative.
Write \(f(x)\) so that \(f(n)=a_n\).
Choose a sufficiently large starting point so every logarithmic factor is defined and the function is positive.
Use increasing positive denominator factors when valid, or determine the sign of \(f'(x)\).
Use a logarithmic substitution and take the upper-endpoint limit.
The integral test is useful here because it turns a delicate comparison of logarithmic terms into a direct question about an improper integral. The hypotheses are part of the argument: a correct integral evaluation does not justify a series conclusion unless the corresponding function also satisfies the test’s conditions on a tail.
Check Your Understanding
For each question, state the relevant hypothesis or substitution before giving a convergence conclusion.
- For \(f(x)=1/[x(\ln x)^{1/2}]\), on what tail is the function positive, continuous, and decreasing?
- Does \(\sum_{n=2}^{\infty}1/[n(\ln n)^2]\) converge or diverge by the integral test?
- What substitution evaluates \(\int dx/[x\ln x(\ln\ln x)^q]\), and what condition on \(q\) gives a finite integral?
- Why can the integral test apply to a series even if its associated function is not decreasing at its first few integer inputs?
- Find an antiderivative of \((\ln x)/x\) and use its limit to classify the associated series.
