5.4 Comparison Tests

Comparison Tests for Series Convergence

Comparison tests let you determine whether a series converges or diverges by relating it to a series whose behavior you already know. Instead of wrestling with a complicated series directly, you compare it to a familiar one (like a p-series or geometric series) and let that known series do the heavy lifting.

Two versions of this test show up constantly: the Direct Comparison Test and the Limit Comparison Test. Direct comparison works through inequalities between terms, while limit comparison works through the ratio of terms. Knowing when to use each one, and picking the right series to compare against, is most of the battle.

Direct Comparison Test

The Direct Comparison Test (DCT) relies on bounding one series with another. If you can trap your series below a convergent one or above a divergent one, you're done.

Here's how it works. Suppose $0 \leq a_n \leq b_n$ for all $n$ beyond some index $N$:

• If $\sum b_n$ converges, then $\sum a_n$ converges. (A series smaller than a convergent series must also converge.)
• If $\sum a_n$ diverges, then $\sum b_n$ diverges. (A series larger than a divergent series must also diverge.)

The tricky part is establishing the inequality. You need to show $a_n \leq b_n$ (or $a_n \geq b_n$) for all sufficiently large $n$. This often involves algebraic manipulation, like noting that $\frac{1}{n^2 + 3} < \frac{1}{n^2}$ because adding 3 to the denominator makes the fraction smaller.

When DCT works well: Series where you can cleanly drop or add terms in the numerator or denominator to create a simpler expression. For example, $\frac{n}{n^3 + 5} < \frac{n}{n^3} = \frac{1}{n^2}$, and since $\sum \frac{1}{n^2}$ converges (p-series with $p = 2$), the original series converges too.

Watch out for direction. If you're trying to prove convergence, your comparison series must sit above the original and converge. If you're trying to prove divergence, it must sit below and diverge. Getting this backwards is a common mistake.

Limit Comparison Test

The Limit Comparison Test (LCT) is your go-to when setting up a clean inequality feels difficult or awkward. Instead of comparing term sizes directly, you compare their ratio as $n \to \infty$.

Given two series $\sum a_n$ and $\sum b_n$ with positive terms, compute:

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

• If $0 < L < \infty$ (a positive, finite number), then both series converge or both diverge.
• If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. (But $L = 0$ with $\sum b_n$ diverging tells you nothing about $\sum a_n$.)
• If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. (But $L = \infty$ with $\sum b_n$ converging tells you nothing about $\sum a_n$.)

The most common and useful case is the first one: you get a finite positive limit, and the two series share the same fate.

Example: Test $\sum \frac{3n^2 + 1}{n^4 - 2n + 7}$. The dominant behavior for large $n$ looks like $\frac{3n^2}{n^4} = \frac{3}{n^2}$, so compare against $b_n = \frac{1}{n^2}$.

$\lim_{n\to\infty} \frac{\frac{3n^2+1}{n^4-2n+7}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{3n^4 + n^2}{n^4 - 2n + 7} = 3$

Since $L = 3$ (positive and finite) and $\sum \frac{1}{n^2}$ converges, the original series converges.

Choosing a Comparison Series

Picking the right comparison series is the most important step. A poor choice leads to inconclusive results or inequalities that point the wrong direction.

p-series $\sum \frac{1}{n^p}$: Converges for $p > 1$, diverges for $p \leq 1$. Use these when the series has polynomial expressions. Look at the highest powers of $n$ in the numerator and denominator, then simplify. For instance, $\frac{n^2 + 1}{n^5 - 3}$ behaves like $\frac{n^2}{n^5} = \frac{1}{n^3}$, so compare to $\frac{1}{n^3}$.

Geometric series $\sum ar^n$: Converges for $|r| < 1$, diverges for $|r| \geq 1$. Use these when the series involves exponential terms like $2^n$, $e^{-n}$, or constant ratios between successive terms. For example, $\frac{1}{3^n + 1} < \frac{1}{3^n} = \left(\frac{1}{3}\right)^n$, which is geometric with $r = \frac{1}{3}$.

General strategy:

1. Identify the dominant terms in the numerator and denominator for large $n$.
2. Simplify the expression using only those dominant terms. This gives your candidate comparison series.
3. Decide whether DCT or LCT is easier. If the inequality is obvious, use DCT. If it's messy, use LCT.
4. For DCT, verify the inequality holds and goes in the right direction (upper bound for convergence, lower bound for divergence).
5. State your conclusion: since the comparison series converges/diverges and the test conditions are met, the original series converges/diverges.

Bounding and Inequalities in Comparison Tests

Setting up valid inequalities is the core skill behind the Direct Comparison Test. A few reliable techniques come up repeatedly.

Making fractions larger (to prove convergence): Shrink the denominator or grow the numerator. Dropping a positive term from the denominator makes the fraction bigger: $\frac{1}{n^2 + n} < \frac{1}{n^2}$. If the bigger series converges, so does the original.

Making fractions smaller (to prove divergence): Grow the denominator or shrink the numerator. For example, $\frac{n}{n^2 + 1} > \frac{n}{n^2 + n^2} = \frac{1}{2n}$. Since $\sum \frac{1}{2n}$ diverges (it's a constant multiple of the harmonic series), the original diverges.

These inequalities only need to hold for all $n$ past some finite index $N$. The first few terms of a series don't affect convergence, so don't worry if the inequality breaks down for small values of $n$.