Why This Growth Comparison Matters
A logarithm grows slowly compared with a positive power of its input. This fact is useful when a sequence or series contains both kinds of factors: it lets us decide which factor controls the long-term size and choose a simpler comparison model.
The comparison is stronger than a visual impression about growth. For any fixed positive exponents, a power of \(n\) eventually exceeds a power of \(\ln n\) by an arbitrarily large factor. The word “eventually” matters: the inequality need not hold for every \(n\), but it does hold for all sufficiently large \(n\).
The Fundamental Limit
The constant \(C\) in the equivalent statement can be any positive number. This follows from the limit: if the ratio tends to zero, then eventually it is less than \(C\). In particular, taking \(C=1\) gives \((\ln n)^\beta<n^\alpha\) for all sufficiently large \(n\).
Why the Theorem Is True
Set \(t=\ln n\). As \(n\) tends to infinity, \(t\) also tends to infinity, and \(n^\alpha=e^{\alpha t}\). The ratio in the theorem can therefore be written as \(t^\beta e^{-\alpha t}\). We show directly that this expression tends to zero.
Let \(T=\max(1,2\beta/\alpha)\). For \(t\ge T\), the function \(F(t)=t^\beta e^{-\alpha t}\) is positive, so its logarithm is defined. Differentiating gives \(\frac{d}{dt}\ln F(t)=\frac{\beta}{t}-\alpha\le-\frac{\alpha}{2}\).
For every \(t\ge T\), the Fundamental Theorem of Calculus gives \(\ln F(t)-\ln F(T)\le-\frac{\alpha}{2}(t-T)\). Exponentiating both sides yields \(0<F(t)\le F(T)e^{-\alpha(t-T)/2}\).
Because \(\alpha>0\), \(e^{-\alpha(t-T)/2}\) tends to zero as \(t\) tends to infinity. The upper bound therefore tends to zero, and \(F(t)\), which is positive and no larger than that bound, also tends to zero.
Finally, substituting \(t=\ln n\) gives \(F(t)= (\ln n)^\beta e^{-\alpha\ln n}=(\ln n)^\beta/n^\alpha\). This proves the stated limit. The proof shows why the result holds: after a sufficiently large input, the exponential factor in \(e^{\alpha t}\) outpaces the fixed power \(t^\beta\).
Turning the Limit into a Comparison Rule
A convenient consequence is that, for any \(\varepsilon>0\) and \(\beta>0\), there is an \(N\) such that $$ (\ln n)^\beta<n^\varepsilon\qquad(n\ge N). $$ The choice of \(\varepsilon\) is ours. That flexibility is what makes the result useful: we can allocate a small part of a power of \(n\) to absorb the logarithm, while leaving the rest for a familiar comparison.
For example, suppose a series has positive terms \((\ln n)^\beta/n^\alpha\). If \(\alpha>1\), choose \(\varepsilon\) with \(0<\varepsilon<\alpha-1\). Eventually, $$ \frac{(\ln n)^\beta}{n^\alpha}<\frac{1}{n^{\alpha-\varepsilon}}. $$ Since \(\alpha-\varepsilon>1\), the comparison series is a convergent \(p\)-series. Direct comparison then proves convergence of the original series.
If \(0<\alpha<1\), then \((\ln n)^\beta\ge1\) for all sufficiently large \(n\). Thus \((\ln n)^\beta/n^\alpha\ge1/n^\alpha\), and the corresponding series diverges by comparison with a divergent \(p\)-series. At the critical exponent \(\alpha=1\), this power comparison alone does not settle every logarithmic case; the logarithmic threshold results are the appropriate tool there.
Worked Examples: Applying the Result
Worked Example: Evaluate a logarithmic-power limit
Find \(\displaystyle\lim_{n\to\infty}\frac{(\ln n)^3}{n^{2/5}}\).
Match the expression with the theorem: \(\beta=3>0\) and \(\alpha=2/5>0\). Therefore, $$ \lim_{n\to\infty}\frac{(\ln n)^3}{n^{2/5}}=0. $$ To check that the theorem applies, both exponents are fixed and positive, as required. The numerator is a fixed logarithmic power and the denominator is a positive power of \(n\), so the ratio has exactly the stated form.
Worked Example: Prove convergence by choosing a comparison power
Determine whether \(\displaystyle\sum_{n=2}^{\infty}\frac{(\ln n)^2}{n^{3/2}}\) converges.
We need a comparison series with exponent greater than \(1\). Choose \(\varepsilon=1/4\). By the theorem, for all sufficiently large \(n\), $$ \frac{(\ln n)^2}{n^{1/4}}<1, \qquad\text{so}\qquad (\ln n)^2<n^{1/4}. $$ For those \(n\), divide this inequality by \(n^{3/2}>0\): $$ 0<\frac{(\ln n)^2}{n^{3/2}} <\frac{n^{1/4}}{n^{3/2}} =\frac{1}{n^{5/4}}. $$ The exponent \(5/4\) is greater than \(1\), so \(\sum 1/n^{5/4}\) converges. The direct comparison test proves that the given series converges. The finitely many terms before the eventual inequality begins do not affect convergence.
Check the comparison by dividing the original term by the proposed bound: the ratio is \((\ln n)^2/n^{1/4}\), which is eventually less than \(1\) by the theorem. The bound therefore has the correct direction.
Worked Example: Prove divergence using a lower bound
Determine whether \(\displaystyle\sum_{n=2}^{\infty}\frac{(\ln n)^2}{n^{1/2}}\) diverges.
For \(n\ge e\), \(\ln n\ge1\), and hence \((\ln n)^2\ge1\). Therefore, for all sufficiently large integers \(n\), $$ \frac{(\ln n)^2}{n^{1/2}}\ge\frac{1}{n^{1/2}}. $$ The series \(\sum 1/n^{1/2}\) diverges because its \(p\)-series exponent \(1/2\) is at most \(1\). By the divergence comparison test, the given series diverges.
Check the direction by dividing the original term by \(1/n^{1/2}\). The ratio is exactly \((\ln n)^2\), which is at least \(1\) once \(n\ge e\). Thus the original terms are indeed no smaller than the divergent comparison terms.
Logarithms in the Denominator
The same theorem helps when the logarithmic factor is in a denominator, but the inequality direction must be handled carefully. For example, if \(\beta>0\) and \(\alpha>1\), then \((\ln n)^\beta\ge1\) eventually, so $$ 0<\frac{1}{n^\alpha(\ln n)^\beta}\le\frac{1}{n^\alpha}. $$ The \(p\)-series comparison immediately proves convergence. If \(0<\alpha<1\), choose \(\varepsilon>0\) so that \(\alpha+\varepsilon<1\). The theorem gives \((\ln n)^\beta<n^\varepsilon\) eventually, and taking reciprocals of positive quantities gives $$ \frac{1}{n^\alpha(\ln n)^\beta}>\frac{1}{n^{\alpha+\varepsilon}}. $$ The comparison series on the right diverges, so the original series diverges as well. Again, this argument does not decide the critical case \(\alpha=1\).
A Common Mistake: Treating “Eventually” as “Always”
The theorem does not say that every positive power of \(n\) exceeds every logarithmic power at every input. For a concrete counterexample, take \(\varepsilon=0.01\) and \(n=e^{100}\). Then $$ \ln n=100, \qquad n^{0.01}=(e^{100})^{0.01}=e\approx2.718. $$ Thus the claim \(\ln n\le n^{0.01}\) is false at this value of \(n\), even though the theorem says it becomes true for all sufficiently large \(n\).
The correction is to keep track of the quantifier: for each fixed positive \(\varepsilon\) and \(\beta\), there exists a threshold \(N\) beyond which \((\ln n)^\beta<n^\varepsilon\). Use the inequality only after that threshold. In a series comparison, the omitted initial segment contains only finitely many terms, so it cannot change whether the series converges or diverges.
Choosing a Useful Model
When a logarithm and a power occur together, first identify the net power of \(n\). Then decide how much of that power can be spent absorbing the logarithm. For a convergence comparison, leave a remaining power \(n^{-p}\) with \(p>1\); for a divergence comparison, seek a lower bound by \(n^{-p}\) with \(p\le1\). The theorem justifies these choices whenever the exponent is not at the critical boundary.
Check Your Understanding
Use the theorem and comparison principles to answer these questions.
- What is the limit of \((\ln n)^7/n^{1/3}\) as \(n\to\infty\), and which exponents make the theorem applicable?
- For \(\sum (\ln n)^4/n^{2}\), choose a positive \(\varepsilon\) that leaves a convergent \(p\)-series as an upper comparison.
- Why does \((\ln n)^2\ge1\) eventually give a useful lower bound for a series with terms \((\ln n)^2/n^{3/4}\)?
- Explain why the theorem does not claim \(\ln n<n^{0.01}\) for every \(n\).
- Why does the power comparison by itself not decide every series with a logarithmic factor at the critical power \(1/n\)?