Comparison Tests

Known series can serve as benchmarks

Many positive-term series do not have partial sums that can be written neatly. Comparison tests avoid that obstacle by asking whether the new terms are eventually smaller or larger than terms from a series whose behavior is already known. The most common benchmarks are geometric series and p-series. The comparison is about the infinite tail, so a finite number of exceptional early terms does not change the conclusion.

The direct comparison test has two useful directions. If $0\le a_n\le b_n$ eventually and $\sum b_n$ converges, then $\sum a_n$ converges. A smaller nonnegative accumulation cannot exceed a finite larger accumulation. If $0\le b_n\le a_n$ eventually and $\sum b_n$ diverges, then $\sum a_n$ diverges. A larger nonnegative accumulation cannot settle when a smaller one already grows without bound.

The direction matters. Showing that a series is smaller than a divergent series says nothing, and showing that it is larger than a convergent series says nothing. These are not annoying exceptions; they follow directly from the logic of the two implications. Before comparing, write the intended benchmark’s behavior and decide whether you need an upper bound for convergence or a lower bound for divergence.

For a direct comparison, the word “eventually” should be made visible in your work. State an index such as $n\ge1$ or $n\ge4$ after which the inequality holds, especially if algebraic rearrangement introduced a restriction. The first finitely many terms can be added, removed, or adjusted without changing convergence, but they still need to be meaningful terms of the original series. This habit keeps a correct tail argument from being applied carelessly at an index where a denominator vanishes or an expression changes sign.

Direct comparison works with inequalities

Consider

$$\sum_{n=1}^{\infty}\frac{1}{n^2+4}.$$

Since $n^2+4\ge n^2$, reciprocal order reverses for positive quantities:

$$0<\frac{1}{n^2+4}\le\frac{1}{n^2}.$$

The p-series $\sum1/n^2$ converges because $p=2>1$. Therefore the given series converges by direct comparison. The extra $4$ in the denominator makes each term smaller, and the inequality makes that intuition mathematically precise.

For divergence, consider $\sum1/(n+1)$. For $n\ge1$, $n+1\le2n$, so

$$\frac{1}{n+1}\ge\frac{1}{2n}=\frac12\cdot\frac1n.$$

The harmonic series diverges, and multiplying it by the positive constant $1/2$ does not change that behavior. Since the new terms are at least this divergent benchmark, the series diverges. The conclusion would be invalid if the inequality pointed the other way, so write the comparison explicitly rather than relying on an impression that denominators are “about the same.”

Limit comparison captures shared leading behavior

Direct inequalities can be awkward when expressions contain several polynomial terms, roots, or rational factors. Limit comparison asks whether two positive-term sequences have the same eventual scale. If

$$\lim_{n\to\infty}\frac{a_n}{b_n}=c$$

for a finite constant $c$ with $0<c<\infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. A positive finite limit says neither sequence eventually dwarfs the other; their terms differ only by an approximately constant factor.

For example, examine

$$\sum_{n=1}^{\infty}\frac{3n+2}{n^3-n+1}.$$

The leading behavior is $3n/n^3=3/n^2$, suggesting comparison with $b_n=1/n^2$. Compute

$$\lim_{n\to\infty} \frac{(3n+2)/(n^3-n+1)}{1/n^2} =\lim_{n\to\infty}\frac{3n^3+2n^2}{n^3-n+1}=3.$$

Because the limit is positive and finite and $\sum1/n^2$ converges, the original series converges. The test did not calculate its sum. It classified the accumulation by its dominant growth terms.

Choosing a benchmark responsibly

For rational expressions in $n$, compare highest powers after simplifying. A numerator growing like $n^p$ and denominator growing like $n^q$ often produces terms behaving like $1/n^{q-p}$. For roots, factor out the highest power inside the root before identifying the dominant part. For exponential expressions, compare ratios to a geometric series, since exponential growth or decay eventually overwhelms polynomial factors.

The phrase “behaves like” is a prediction, not a proof. Direct comparison turns it into an inequality; limit comparison turns it into a computed ratio limit. Do not omit the positive-term requirement. Alternating series use different logic because cancellation can radically alter the accumulation. A comparison of absolute values may later be useful, but it answers absolute convergence rather than the original signed series by default.

There is also a practical reason to prefer the simplest valid comparison. An exact algebraic inequality may be faster than a lengthy ratio limit when denominators visibly order themselves. Conversely, a limit comparison may be clearer than forcing a chain of estimates through nested radicals. The goal is not to demonstrate the most complicated tool available. The goal is to make the shared long-run behavior transparent enough that another reader can audit the conclusion.

Retrieval check

Classify $\sum_{n=1}^{\infty}(n+5)/(n^3+2)$ using limit comparison, and classify $\sum_{n=1}^{\infty}1/(n^2+n)$ using direct comparison. In each case name the benchmark, show either an inequality or the ratio limit, and state the benchmark’s known behavior. Then explain why comparing the second series to $1/n$ from above would not prove convergence. The next lesson introduces alternating signs and asks when cancellation produces conditional versus absolute convergence.