Why Bound Logarithms by Powers?
A logarithm grows more slowly than any positive power of its input. This fact makes logarithmic factors manageable in many series: an expression involving \(\ln n\) can often be bounded by a constant times a power of \(n\), after which a p-series comparison decides convergence or divergence.
The exponent in the bound matters. A bound that is valid but too large may lead to a power series that does not answer the question. The goal is to choose a small enough power when proving convergence, and a suitable upper bound on the logarithm when constructing a lower comparison for divergence.
An Elementary Bound for the Logarithm
For \(x\geq 1\), the natural logarithm has the integral representation \(\ln x=\int_1^x 1/t\,dt\). If \(a>0\) and \(t\geq 1\), then \(t^a\geq 1\), so \(1/t\leq t^{a-1}\). Integrating this inequality gives a useful bound.
This bound holds for every \(a>0\) and \(x\geq 1\). In particular, it implies that for any \(b>0\), \(\ln x\leq x^b\) once \(x\) is sufficiently large: apply the displayed inequality with exponent \(b/2\), then take \(x\) large enough that \(2x^{b/2}/b\leq x^b\). Thus the familiar bound \(\ln n\leq n^b\) is an eventual bound, while the bound with a constant is available directly.
How to Choose the Exponent
Suppose a positive series has terms \((\ln n)^r/n^p\), where \(r>0\). Applying the bound with exponent \(a\) gives
For convergence, choose \(a>0\) so that \(p-ar>1\). Such a choice is possible whenever \(p>1\). The resulting upper comparison is a convergent p-series, so the original positive series converges by direct comparison.
For divergence in a denominator-log example, the same upper bound on the logarithm works in the opposite direction after taking reciprocals. For \(r>0\),
If \(p<1\), choose \(a>0\) small enough that \(p+ar\leq 1\). The lower comparison then diverges, proving divergence of the original series. The comparison direction is essential: divergence follows from a lower bound by a divergent positive series, not from an upper bound by one.
Worked Comparisons
Worked Example: A Logarithm in the Numerator
Determine whether \(\sum_{n=2}^{\infty}\ln n/n^{5/4}\) converges. Use \(a=1/8\). The elementary bound gives \(\ln n\leq 8n^{1/8}\), since \(1/a=8\). Therefore, for every \(n\geq 2\),
The comparison series \(\sum 8/n^{9/8}\) converges because \(9/8>1\). The terms are positive, and the original terms are no larger than those of this convergent series. Direct comparison proves that \(\sum_{n=2}^{\infty}\ln n/n^{5/4}\) converges.
Check the exponent calculation: \(5/4=10/8\), so \(5/4-1/8=9/8\). The selected exponent therefore does give a convergent p-series.
Worked Example: A Squared Logarithm
Determine whether \(\sum_{n=2}^{\infty}(\ln n)^2/n^{3/2}\) converges. Again choose \(a=1/8\), so \(\ln n\leq 8n^{1/8}\). Both sides are nonnegative, and squaring preserves the inequality:
It follows that
Here \(3/2=6/4\), so \(3/2-1/4=5/4\). Since \(5/4>1\), the comparison p-series converges. Direct comparison proves convergence of the given series. Substituting the chosen bound back into the original term produces exactly the displayed upper bound, including the constant \(64\).
Worked Example: A Logarithm in the Denominator
Determine whether \(\sum_{n=2}^{\infty}1/[n^{3/4}(\ln n)^2]\) diverges. Use \(a=1/16\), for which \(\ln n\leq 16n^{1/16}\). Since \(n\geq2\), \(\ln n>0\), so we may square the inequality and take reciprocals:
The exponent is \(3/4+1/8=6/8+1/8=7/8\). The comparison series \(\sum 1/(256n^{7/8})\) diverges because \(7/8<1\). Every original term is at least the corresponding comparison term, so the original positive series diverges by direct comparison. The reciprocal step is valid because both quantities being compared are positive.
A General Conclusion and Its Limits
These arguments give a useful summary for powers of logarithms. Let \(r>0\), and consider terms with \(n\geq2\). If \(p>1\), then \((\ln n)^r/n^p\) gives a convergent series: choose \(a>0\) with \(p-ar>1\), and apply the upper comparison. If \(p\leq1\), the series with numerator \((\ln n)^r\) diverges. Indeed, for all sufficiently large \(n\), \(\ln n\geq1\), so its terms are at least \(1/n^p\), whose series diverges when \(p\leq1\).
For denominator logarithms, if \(p<1\), the series with terms \(1/[n^p(\ln n)^r]\) diverges: choose \(a>0\) so small that \(p+ar<1\), and use the lower comparison above. If \(p>1\), it converges, because eventually \(\ln n\geq1\), giving \(1/[n^p(\ln n)^r]\leq1/n^p\).
These power bounds alone do not decide every case with \(p=1\) and a logarithm in the denominator. That is a borderline logarithmic case, addressed by the earlier material on logarithmic factors in harmonic-type series. Recognizing the limit of a comparison method is part of using it correctly.
A Common Comparison Error
A frequent mistake is to use an upper bound by a divergent series to claim divergence. For example, for \(n\geq2\), \(\ln n/n^{3/2}\leq 1/n\) eventually, but \(\sum 1/n\) diverges. That inequality does not determine whether the smaller series converges or diverges: a series of positive terms can be smaller than a divergent series and still converge.
For this particular series, use the logarithm bound instead. With \(a=1/4\), \(\ln n\leq4n^{1/4}\), and hence
Since \(\sum 4/n^{5/4}\) converges, direct comparison proves that \(\sum \ln n/n^{3/2}\) converges. The correction is to choose a comparison inequality whose direction and convergence behavior support the conclusion.
Why the Method Is Useful
Logarithmic factors can make a term look substantially different from a pure power, but their growth is slow enough that a small increase or decrease in the power of \(n\) can control them. The integral inequality explains this directly: on \(t\geq1\), the function \(t^{a-1}\) eventually provides an upper bound for \(1/t\), and its integral is a power expression. Raising the resulting logarithm bound to a positive power preserves a manageable power comparison.
In practice, first identify whether a logarithm is in the numerator or denominator. Then decide whether the proof needs an upper or a lower bound on the terms. Finally, choose \(a\) so that the resulting power exponent lies on the appropriate side of \(1\). This keeps the argument explicit and prevents a comparison that is correct but inconclusive.
Check Your Understanding
For each question, identify the comparison direction and the power exponent before deciding what the bound proves.
- For which values of \(a\) does the inequality \(\ln x\leq (x^a-1)/a\) hold when \(x\geq1\)?
- Use an upper bound for \(\ln n\) to compare \(\sum_{n=2}^{\infty}(\ln n)^3/n^2\) with a p-series.
- Why does an upper bound on \(\ln n\) give a lower bound for \(1/[n^p(\ln n)^r]\), when \(r>0\)?
- What condition on \(a\) makes the comparison in \(\sum(\ln n)^r/n^p\) a convergent p-series?
- Why does \(\sum a_n\leq\sum 1/n\), with positive terms, not prove that \(\sum a_n\) diverges?
