Mixed Power and Logarithm Comparison Models

Introduction

Many positive series have terms that combine a power of the index with a power of its logarithm. The central task is to identify the right benchmark: the power of $n$ determines the main threshold, while the logarithm decides the outcome when that power is exactly critical.

Fixed multipliers and fixed shifts can make a formula look different from a familiar model. They do not usually change the convergence classification, but they must be handled through a valid comparison rather than ignored without explanation.

The Power-and-Logarithm Model

For $n\ge 2$, consider terms of the form $n^{-p}(\ln n)^{-q}$, where $p$ and $q$ are real numbers. Equivalently, these terms can be written as $1/[n^p(\ln n)^q]$. The logarithm is positive for $n\ge2$, so these terms are positive for every real $p$ and $q$.

Definition: A power-and-logarithm model is a sequence whose terms, for all sufficiently large $n$, have the form $C n^{-p}(\ln n)^{-q}$, or are asymptotically equivalent to such terms, where $C>0$ is fixed and $p,q\in\mathbb R$. Asymptotic equivalence means that the ratio of the two positive sequences tends to $1$.

For this model, the classification is governed by two thresholds. If $p>1$, the series converges for every fixed $q$. If $p<1$, it diverges for every fixed $q$. When $p=1$, the logarithmic exponent matters: the series converges exactly when $q>1$, and diverges when $q\le1$. The case $p=1,q=1$ is the harmonic logarithmic threshold.

Theorem: For real $p,q$, the positive-term series $$ \sum_{n=2}^{\infty}\frac{1}{n^p(\ln n)^q} $$ converges if $p>1$, diverges if $p<1$, and, if $p=1$, converges exactly when $q>1$.

The powers of $n$ set the primary threshold at $p=1$. Away from that boundary, any fixed power of $\ln n$ is too small to change the classification. At the boundary, the logarithmic factor becomes decisive.

Why Equivalent Models Have the Same Classification

A useful comparison works when the ratio of two positive terms approaches a finite, strictly positive number. Suppose $a_n,b_n>0$ for all sufficiently large $n$, and $a_n/b_n\to L$ with $0<L<\infty$. Then, for all sufficiently large $n$, the ratio lies between $L/2$ and $3L/2$. Multiplying by $b_n$ gives bounds on $a_n$, so the partial sums of the two series are bounded by constant multiples of one another. Consequently, $\sum a_n$ and $\sum b_n$ either both converge or both diverge. This is the limit comparison principle.

Here is why fixed shifts and multipliers often lead to such a ratio. Let $A>0$ and let $a,b,c,d$ be fixed real numbers, with the expressions defined and positive for all sufficiently large $n$. For fixed real $p,q$, compare

t_n=\frac{A}{(an+b)^p[\ln(cn+d)]^q}

with $n^{-p}(\ln n)^{-q}$, assuming $a>0$ and $c>0$. Their ratio is

\frac{t_n}{n^{-p}(\ln n)^{-q}} = A\left(\frac{n}{an+b}\right)^p \left(\frac{\ln n}{\ln(cn+d)}\right)^q.

Since $n/(an+b)\to1/a$, and $\ln(cn+d)/\ln n\to1$, this ratio tends to $A a^{-p}$, a positive finite constant. The two series therefore have the same classification. This calculation explains why fixed shifts inside a power or logarithm, and fixed positive multipliers, do not alter the benchmark’s convergence behavior.

Choosing a Benchmark

For a mixed term, first identify the power of $n$ and the power of the logarithm. If the term has a denominator $(\ln n)^q$, then $q$ is positive when the logarithm is in the denominator and negative when the logarithm is in the numerator. Next, compare the power exponent with $1$. Only when it equals $1$ do you need the logarithmic threshold $q=1$.

Sometimes the term contains a logarithmic factor and a direct comparison with $n^{-p}$ has ratio tending to zero or infinity, not to a finite positive constant. In that situation, choose a nearby power benchmark. The fact that powers dominate logarithms gives the required direction: for any $\varepsilon>0$, $n^\varepsilon$ eventually grows faster than any fixed power of $\ln n$. Thus a logarithm can be absorbed by a small change in the power exponent.

Read the power of $n$.
Write the principal factor as $n^{-p}$ and compare $p$ with $1$.

Check the logarithm if needed.
If $p=1$, write the logarithmic factor as $(\ln n)^{-q}$ and compare $q$ with $1$.

Account for fixed shifts and constants.
Use a ratio limit to show that the given terms are comparable to the canonical model.

Use a nearby power if the power is not critical.
When a logarithm prevents a finite positive ratio with the first benchmark, shift the benchmark exponent slightly and compare in the needed direction.

Worked Examples

Worked Example: A Subcritical Power with a Strong Logarithmic Denominator

Classify the positive-term series

\sum_{n=2}^{\infty}\frac{4}{n^{3/4}(\ln n)^5}.

The power exponent is $p=3/4<1$. The model criterion says that any fixed logarithmic power leaves a series with this subcritical power divergent. A direct ratio with $n^{-3/4}$ tends to zero, so instead compare with the divergent benchmark $n^{-7/8}$, whose power exponent is still below $1$.

\frac{4}{n^{3/4}(\ln n)^5}\bigg/n^{-7/8} = \frac{4n^{1/8}}{(\ln n)^5} \longrightarrow \infty.

The limit follows because the positive power $n^{1/8}$ grows faster than the fixed logarithmic power $(\ln n)^5$. Therefore the displayed ratio is eventually at least $1$, so the given term is eventually at least $n^{-7/8}$. Since $\sum n^{-7/8}$ diverges, direct comparison proves that the given series diverges.

Check: the comparison is in the required direction: the given terms are eventually no smaller than the divergent benchmark. The conclusion is divergence, despite the fifth power of the logarithm in the denominator.

Worked Example: A Convergent Logarithmic Boundary Case

Classify

\sum_{n=2}^{\infty} \frac{6}{(2n+1)[\ln(n+1)]^{3/2}}.

The shifts suggest the benchmark $1/[n(\ln n)^{3/2}]$. Compute the ratio explicitly:

\frac{\dfrac{6}{(2n+1)[\ln(n+1)]^{3/2}}} {\dfrac{1}{n(\ln n)^{3/2}}} = \frac{6n}{2n+1} \left(\frac{\ln n}{\ln(n+1)}\right)^{3/2}.

As $n\to\infty$, the first factor tends to $3$, because $6n/(2n+1)\to3$. Also, $\ln(n+1)/\ln n\to1$, so the second factor tends to $1$. The full ratio therefore tends to $3$, which is finite and strictly positive. By limit comparison, the given series has the same classification as $\sum 1/[n(\ln n)^{3/2}]$. Here $p=1$ and $q=3/2>1$, so that benchmark converges.

Check: the ratio limit is $3$, not $1$, but that is sufficient for limit comparison. The given series converges.

Worked Example: The Divergent Logarithmic Threshold

Classify

\sum_{n=2}^{\infty}\frac{8}{(4n+7)\ln(2n+3)}.

Use the critical benchmark $1/(n\ln n)$. The ratio is

\frac{\dfrac{8}{(4n+7)\ln(2n+3)}}{\dfrac{1}{n\ln n}} = \frac{8n\ln n}{(4n+7)\ln(2n+3)} = \frac{8n}{4n+7}\cdot\frac{\ln n}{\ln(2n+3)}.

The first factor tends to $2$, since $8n/(4n+7)\to2$. The logarithm ratio tends to $1$, since $\ln(2n+3)/\ln n\to1$. Thus the complete ratio tends to $2$, a finite positive limit. The benchmark has $p=1,q=1$, which is the divergent boundary case. Limit comparison proves that the given series diverges as well.

Check: both the shifted linear factor and the shifted logarithm contribute fixed limiting ratios; their product is $2$. The benchmark diverges, so the given series diverges.

Worked Example: A Supercritical Power with a Logarithmic Numerator

Classify

\sum_{n=2}^{\infty}\frac{3(\ln(n+2))^4}{n^{5/4}}.

The logarithm is in the numerator, so it makes the terms larger than $n^{-5/4}$. To obtain a convergent upper benchmark, compare with $n^{-9/8}$, whose exponent is still greater than $1$:

\frac{\dfrac{3(\ln(n+2))^4}{n^{5/4}}}{n^{-9/8}} = \frac{3(\ln(n+2))^4}{n^{1/8}} \longrightarrow 0.

The limit is zero because $n^{1/8}$ grows faster than the fixed logarithmic power. Therefore, for all sufficiently large $n$, the given term is at most $n^{-9/8}$. The benchmark series converges because $9/8>1$; direct comparison proves that the given series converges.

Check: the ratio tends to zero, so the given terms are eventually smaller than the convergent benchmark. The conclusion is convergence.

A Common Mistake: Treating Every Logarithmic Denominator as Decisive

A logarithmic denominator does not automatically make a series converge. For example, it is incorrect to argue that $\sum 1/[n^{1/2}(\ln n)^{10}]$ converges merely because $(\ln n)^{10}$ grows without bound. Here the power exponent is $1/2<1$, so the power threshold decides the classification: the series diverges.

A comparison makes the reason concrete. Divide the term by the divergent benchmark $n^{-3/4}$:

\frac{\dfrac{1}{n^{1/2}(\ln n)^{10}}}{n^{-3/4}} = \frac{n^{1/4}}{(\ln n)^{10}} \longrightarrow\infty.

Thus the terms are eventually at least $n^{-3/4}$, and the series diverges by direct comparison. The correction is to inspect the power exponent first; logarithmic factors determine the outcome only at the critical power $p=1$.

A related bookkeeping error is to confuse the exponent $q$ with the exponent appearing in the denominator. In the form $n^{-p}(\ln n)^{-q}$, the series $1/(n\ln n)$ has $q=1$, not $q=-1$. Its logarithmic exponent is exactly critical, and the series diverges.

Why These Models Matter

Power-and-logarithm terms occur naturally after simplifying rational expressions, radicals, and products involving logarithms. Recognizing their benchmark lets you classify a series without needing its partial sums or an exact closed form. The method also separates two different questions: fixed constants and shifts affect the limiting comparison ratio, while the exponents determine whether the benchmark converges.

For reliable decisions, write the model explicitly before classifying it. In particular, record whether the logarithm is in the numerator or denominator, state the resulting value of $q$, and use a ratio or direct inequality whose direction supports the conclusion.

Check Your Understanding

For each question, identify a suitable power-and-logarithm benchmark and state the convergence classification.

Classify $\sum_{n=2}^{\infty}1/[n^{6/5}(\ln n)^{-3}]$.
Classify $\sum_{n=2}^{\infty}1/[n(\ln n)^{4/5}]$.
What is the limit of the ratio of $5/[(3n+2)(\ln(n+4))^2]$ to $1/[n(\ln n)^2]$, and what does it imply?
Why does a logarithmic denominator with a fixed positive exponent not make $\sum n^{-2/3}(\ln n)^{-7}$ converge?
For a model $n^{-p}(\ln n)^{-q}$ with $p=1$, what condition on $q$ gives convergence?

Mixed Power and Logarithm Comparison Models

What You'll Learn

Introduction

The Power-and-Logarithm Model

Why Equivalent Models Have the Same Classification

Choosing a Benchmark

Worked Examples

Worked Example: A Subcritical Power with a Strong Logarithmic Denominator

Worked Example: A Convergent Logarithmic Boundary Case

Worked Example: The Divergent Logarithmic Threshold

Worked Example: A Supercritical Power with a Logarithmic Numerator

A Common Mistake: Treating Every Logarithmic Denominator as Decisive

Why These Models Matter

Check Your Understanding

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