Comparison Tests for Series 📘

students, imagine you are trying to decide whether a giant stack of boxes will fit in a room. Instead of measuring every box exactly, you compare the stack to another stack you already understand. That is the big idea behind comparison tests in Calculus 2. In this lesson, you will learn how to decide whether an infinite series converges or diverges by comparing it to a simpler series whose behavior is already known. This is a powerful tool in Convergence Tests I, right alongside the divergence test, the integral test, and $p$-series.

What comparison tests are trying to do

A series is an expression like $\sum_{n=1}^{\infty} a_n$, where you add infinitely many terms. The main question is: does the sum approach a finite value, or does it grow without bound? Comparison tests answer that question by using inequalities.

The logic is simple:

If $a_n$ is always smaller than a series that is known to converge, then $a_n$ may also converge.
If $a_n$ is always larger than a series that is known to diverge, then $a_n$ may also diverge.

This works best when the terms are nonnegative, because inequalities are easier to interpret when every term is $\ge 0$.

There are two main comparison tests you should know:

the direct comparison test
the limit comparison test

Both belong to the same family of ideas: compare a hard series to an easier one. ✅

Direct comparison test: the basic strategy

The direct comparison test uses inequalities between terms.

Suppose $0 \le a_n \le b_n$ for all sufficiently large $n$.

Then:

if $\sum b_n$ converges, $\sum a_n$ also converges
if $\sum a_n$ diverges, $\sum b_n$ also diverges

This test is especially useful when you can sandwich one series against another that you already recognize, such as a geometric series or a $p$-series.

Why this makes sense

If every term of $\sum a_n$ is smaller than the matching term of a convergent series $\sum b_n$, then $\sum a_n$ cannot “add up” to something bigger enough to diverge. On the other hand, if $\sum a_n$ is already too large to converge, then any larger series $\sum b_n$ must also diverge.

Example 1: comparing to a $p$-series

Consider

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

We know that for every $n \ge 1$,

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

and

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

is a $p$-series with $p=2$, so it converges.

Since $0 \le \frac{1}{n^2+1} \le \frac{1}{n^2}$ and the larger series converges, the direct comparison test tells us

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

also converges.

Example 2: proving divergence by comparison

Consider

$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+1}$$

For large $n$,

$$\frac{1}{\sqrt{n}+1} > \frac{1}{2\sqrt{n}}$$

because $\sqrt{n}+1 < 2\sqrt{n}$ when $n>1$.

Now

$$\sum_{n=1}^{\infty} \frac{1}{2\sqrt{n}} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$$

and this is a $p$-series with $p=\frac{1}{2}$, so it diverges.

Since $\frac{1}{\sqrt{n}+1}$ is greater than a divergent series term-by-term, the direct comparison test shows

$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+1}$$

diverges.

How to choose the right comparison series

students, the hardest part is often not the test itself but choosing the comparison series. A good comparison series usually comes from one of these familiar types:

Geometric series: $\sum ar^n$ or $\sum ar^{n-1}$
$p$-series: $\sum \frac{1}{n^p}$
A simpler algebraic expression with the same long-term behavior

A good rule is to focus on what happens as $n \to \infty$. For large $n$, smaller terms like constants often become unimportant compared with the highest powers of $n$.

Example 3: rational function comparison

Look at

$$\sum_{n=1}^{\infty} \frac{5n+1}{n^3+4}$$

For large $n$,

$$\frac{5n+1}{n^3+4} \approx \frac{5n}{n^3} = \frac{5}{n^2}$$

To make this rigorous, note that for $n \ge 1$,

$$5n+1 \le 6n$$

and

$$n^3+4 \ge n^3$$

$$0 \le \frac{5n+1}{n^3+4} \le \frac{6n}{n^3} = \frac{6}{n^2}$$

Since

$$\sum_{n=1}^{\infty} \frac{6}{n^2}$$

converges by the $p$-series test, the given series converges by direct comparison.

Real-world picture

Think of each term as a tiny amount of water filling a cup. If the water in your cup is always less than the amount in another cup that is known to hold a finite amount, then your cup also has only a finite amount. That is the intuition behind convergence by comparison 💧

Limit comparison test: when direct inequalities are hard

Sometimes it is difficult to prove a clean inequality like $a_n \le b_n$, even though the terms clearly behave similarly. That is when the limit comparison test becomes useful.

If $a_n > 0$ and $b_n > 0$, and

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

where $0 < c < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Why it works

If the ratio of the terms approaches a positive finite number, then the two series have the same long-term size. They grow at essentially the same rate, so they share the same convergence behavior.

Example 4: using limit comparison

Consider

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

A natural comparison is

$$\sum_{n=1}^{\infty} \frac{1}{n}$$

because the leading terms suggest the terms behave like $\frac{2n^2}{n^3} = \frac{2}{n}$.

Let

$$a_n = \frac{2n^2+3}{n^3-1}, \quad b_n = \frac{1}{n}$$

Then

$$\frac{a_n}{b_n} = n\cdot \frac{2n^2+3}{n^3-1} = \frac{2n^3+3n}{n^3-1}$$

Now compute the limit:

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

Since the limit is a positive finite number, $\sum a_n$ and $\sum b_n$ behave the same way. But

$$\sum_{n=1}^{\infty} \frac{1}{n}$$

is the harmonic series, and it diverges. Therefore,

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

diverges too.

Common mistakes and how to avoid them

students, here are some common errors students make:

Comparing a series with terms that are not all nonnegative
Choosing the wrong comparison series
Forgetting that the comparison must work for all sufficiently large $n$, not necessarily every single term
Using direct comparison when limit comparison is easier
Assuming that if $a_n \le b_n$ and $\sum b_n$ diverges, then $\sum a_n$ must diverge. That conclusion is not valid.

For example, if $0 \le a_n \le b_n$ and $\sum b_n$ diverges, you cannot conclude anything definite about $\sum a_n$ from that alone. The smaller series might converge or diverge.

This is why it is important to know exactly what each test can and cannot prove.

How comparison tests fit into Convergence Tests I

Comparison tests are part of a larger toolkit for deciding whether series converge. In Convergence Tests I, you commonly study:

the divergence test, which checks whether $\lim_{n\to\infty} a_n = 0$
$p$-series, which give a standard class of convergent and divergent examples
the integral test, which compares a series to an improper integral
comparison tests, which compare one series directly or indirectly to another known series

Together, these tests help you analyze many series that do not have a simple closed-form sum. Comparison tests are especially useful when terms look like fractions, rational functions, or expressions that resemble a known series after simplifying the highest powers.

A strong Calculus 2 problem-solving habit is to ask:

What does this term look like for large $n$?
Which familiar series is it close to?
Can I prove a direct inequality, or is limit comparison easier?

Conclusion

Comparison tests give you a practical way to study infinite series by using known examples as benchmarks. The direct comparison test uses inequalities to show that one series is smaller or larger than another. The limit comparison test checks whether two positive series have the same long-term growth rate. Both methods connect naturally to $p$-series, geometric series, the divergence test, and the integral test. students, once you learn to recognize patterns and choose a good comparison series, many difficult convergence problems become much more manageable. 🎯

Study Notes

Comparison tests help determine whether $\sum a_n$ converges or diverges by comparing it to a known series.
The direct comparison test works with inequalities like $0 \le a_n \le b_n$.
If $\sum b_n$ converges and $0 \le a_n \le b_n$, then $\sum a_n$ converges.
If $\sum a_n$ diverges and $0 \le a_n \le b_n$, then $\sum b_n$ diverges.
The limit comparison test uses $\lim_{n\to\infty} \frac{a_n}{b_n} = c$ with $0 < c < \infty$.
If that limit exists, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.
Common comparison series include geometric series and $p$-series.
For large $n$, the highest powers usually control the behavior of rational expressions.
Comparison tests are part of Convergence Tests I, along with the divergence test, integral test, and $p$-series.
The most important skill is choosing a comparison series whose behavior is already known.