Limit Comparison Test: Statement, Rules & Examples

What Is The Limit Comparison Test?

The Limit Comparison Test is a rule for deciding the convergence or divergence of a positive-term series $\sum a_n$ by comparing it against a benchmark series $\sum b_n$ whose behaviour is already settled. Instead of proving a term-by-term inequality, you compute a single limit of the ratio of the terms.

State it precisely. Suppose $a_n > 0$ and $b_n > 0$ for all $n$ (at least from some point on), and form the limit of their ratio:

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

$$\text{If } 0 < L < \infty, \text{ then } \sum_{n=1}^{\infty} a_n \text{ and } \sum_{n=1}^{\infty} b_n \text{ both converge or both diverge.}$$

The hypotheses matter. Both series must have positive terms, and $L$ must come out to a finite number that is strictly greater than zero. When that happens, the two series are locked together: knowing the fate of the simple comparison series $\sum b_n$ hands you the fate of $\sum a_n$ for free. The whole method sits on top of the convergence and divergence of series, because the benchmark you compare against is always a series whose convergence you have already established.

How Do You Choose The Comparison Series?

The skill in the Limit Comparison Test is picking $b_n$, and there is a reliable rule: keep only the dominant powers of the numerator and denominator of $a_n$, and throw everything else away. For large $n$, the highest power in each part is all that matters, so the leftover is the series your terms actually behave like.

• If $a_n = \dfrac{2n^2 + 3}{n^3 + 5}$, the numerator behaves like $2n^2$ and the denominator like $n^3$, so $a_n$ behaves like $\dfrac{2n^2}{n^3} = \dfrac{2}{n}$. Drop the constant and take $b_n = \dfrac{1}{n}$.

• If $a_n = \dfrac{1}{\sqrt{n^2 + 1}}$, the denominator behaves like $\sqrt{n^2} = n$, so take $b_n = \dfrac{1}{n}$.

• If $a_n = \dfrac{1}{n^2 - 1}$, the denominator behaves like $n^2$, so take $b_n = \dfrac{1}{n^2}$.

The benchmark $\sum b_n$ that falls out is almost always a p-series $\sum \frac{1}{n^p}$ or a geometric series $\sum ar^n$, precisely because those are the two families whose convergence you can read off instantly. A p-series converges exactly when $p > 1$; a geometric series converges exactly when $|r| < 1$.

How Do You Apply The Limit Comparison Test? (Worked Examples)

Each example follows the same three moves: choose $b_n$ by dominant powers, compute $L = \lim_{n \to \infty} \frac{a_n}{b_n}$, then read off the shared fate.

Example 1: A series that converges.

Test $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - 1}$.

The denominator behaves like $n^2$, so take $b_n = \dfrac{1}{n^2}$, a p-series with $p = 2 > 1$ (convergent). Form the ratio:

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

$$L = \lim_{n \to \infty} \frac{1}{1 - \frac{1}{n^2}} = \frac{1}{1 - 0} = 1$$

Since $0 < L = 1 < \infty$ and $\sum \frac{1}{n^2}$ converges, both series converge.

Final answer: the series converges.

Example 2: A series that diverges.

Test $\displaystyle\sum_{n=1}^{\infty} \frac{2n^2 + 3}{n^3 + 5}$.

The terms behave like $\frac{2n^2}{n^3} = \frac{2}{n}$, so take $b_n = \dfrac{1}{n}$, the harmonic series (divergent). Form the ratio:

$$\frac{a_n}{b_n} = \frac{\frac{2n^2 + 3}{n^3 + 5}}{\frac{1}{n}} = \frac{n(2n^2 + 3)}{n^3 + 5} = \frac{2n^3 + 3n}{n^3 + 5}$$

Divide numerator and denominator by $n^3$:

$$L = \lim_{n \to \infty} \frac{2 + \frac{3}{n^2}}{1 + \frac{5}{n^3}} = \frac{2 + 0}{1 + 0} = 2$$

Since $0 < L = 2 < \infty$ and $\sum \frac{1}{n}$ diverges, both series diverge.

Final answer: the series diverges.

Example 3: A square root in the denominator.

Test $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2 + 1}}$.

The denominator behaves like $\sqrt{n^2} = n$, so take $b_n = \dfrac{1}{n}$ (divergent). Form the ratio:

$$\frac{a_n}{b_n} = \frac{\frac{1}{\sqrt{n^2 + 1}}}{\frac{1}{n}} = \frac{n}{\sqrt{n^2 + 1}} = \frac{1}{\sqrt{1 + \frac{1}{n^2}}}$$

$$L = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^2}}} = \frac{1}{\sqrt{1 + 0}} = 1$$

Since $0 < L = 1 < \infty$ and $\sum \frac{1}{n}$ diverges, both series diverge.

Final answer: the series diverges.

What Happens When The Limit Is 0 Or Infinity?

The clean statement needs $0 < L < \infty$. The two boundary cases still say something, but each works in only one direction, and reading them backwards is where marks are lost.

• If $L = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. A limit of $0$ means $a_n$ is eventually much smaller than $b_n$, so being dominated by a convergent series forces convergence. This tells you nothing when $\sum b_n$ diverges.

• If $L = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges. A limit of infinity means $a_n$ eventually dwarfs $b_n$, so sitting above a divergent series forces divergence. This tells you nothing when $\sum b_n$ converges.

The pattern is worth saying out loud. The $L = 0$ case is only useful with a convergent benchmark, and the $L = \infty$ case is only useful with a divergent benchmark. Pair either one the wrong way and the test is silent, which is exactly the moment to pick a better $b_n$.

Table: How the value of the limit $L$ is read.

How Is It Different From The Direct Comparison Test?

Both tests compare $\sum a_n$ to a known series, but they ask for different work. The comparison test needs a genuine term-by-term inequality: to prove convergence you must show $a_n \le b_n$ for a convergent $\sum b_n$, and to prove divergence you must show $a_n \ge b_n$ for a divergent one. The Limit Comparison Test replaces that inequality with a single limit.

The difference shows up on Example 1. Compare $\sum \frac{1}{n^2 - 1}$ to $\sum \frac{1}{n^2}$:

$$\frac{1}{n^2 - 1} > \frac{1}{n^2}$$

The inequality runs the wrong way for a direct comparison. Your terms sit above a convergent series, and a term sitting above a convergent series proves nothing. To rescue the direct comparison you would have to find a cleverer bound, such as $\frac{1}{n^2 - 1} \le \frac{2}{n^2}$ for $n \ge 2$, and justify the constant. The Limit Comparison Test skips all of that: the ratio tends to $1$, so the two series simply share a fate, which is why it is the test to reach for when a clean inequality is awkward to produce.

Which Series Tests Work Alongside It?

The Limit Comparison Test is one tool in the standard convergence toolkit, and it pairs naturally with the others. Each has a shape of series it was built for.

Table: Where the Limit Comparison Test sits among the convergence tests.

A workable habit: run the divergence test first to rule out obvious divergence, and if the terms are a ratio of polynomials or roots, reach straight for the Limit Comparison Test with the dominant-power benchmark. Save the ratio test and root test for terms built from powers or an $n$ factorial.

Why Does The Limit Comparison Test Work?

The idea is that a finite positive limit means the two sets of terms stay in fixed proportion, and a fixed proportion cannot change whether a sum is finite.

• A finite positive $L$ traps $a_n$ between two multiples of $b_n$. If $\frac{a_n}{b_n} \to L$ with $L > 0$, then for large $n$ the ratio sits near $L$, so $a_n$ lies between, say, $\frac{L}{2} b_n$ and $2L b_n$. The terms are sandwiched between constant multiples of $b_n$.

• Scaling a series by a positive constant never changes its fate. Multiplying every term of a convergent series by a fixed number keeps it convergent, and the same holds for divergence. So if $b_n$ is squeezed by constants around $a_n$, the two series must rise or fall together.

• The dominant term is doing all the work. Choosing $b_n$ by the highest powers is choosing the part of $a_n$ that survives in the limit. Everything you dropped, the $+3$ and the $-1$ and the $+5$, becomes negligible next to the leading power as $n$ grows.

Seen this way, the test is a formal version of a plain idea: if your terms are eventually a near-constant multiple of terms you understand, your sum behaves like a sum you understand.

Who Shaped The Limit Comparison Test?

The comparison of series to known benchmarks is old; sharpening it into a clean limit statement came with the drive for rigour in the 1800s.

Two figures bracket the story:

• Augustin-Louis Cauchy (1789–1857, France) gave the comparison principle its first rigorous form in 1821, tying convergence to the behaviour of partial sums and to comparison with known series.

• Alfred Pringsheim (1850–1941, Germany) systematised the convergence tests later in the century, organising the comparison, ratio, and root tests into the toolkit taught today.

Where Is The Limit Comparison Test Used In The Real World?

Deciding whether an infinite sum settles is a reliability check that runs under a good deal of applied mathematics, and the Limit Comparison Test is often the quickest way to make that call.

• Signal processing. Fourier coefficients of a signal form a series, and an engineer checks its convergence by comparing the coefficients to a p-series before trusting a reconstructed waveform.

• Physics. Perturbation series in quantum mechanics and electromagnetism are tested for convergence term-by-shape against known benchmarks before a truncated sum is used as an approximation.

• Numerical computing. Deciding how many terms of a slowly converging series a library must add for a target accuracy starts with knowing the series converges at all, which a limit comparison settles fast.

• Probability. Expected values and generating functions are infinite sums, and comparing their terms to a geometric or p-series decides whether the expectation is a finite number.

• Economics. Discounted infinite streams of payments are compared to geometric benchmarks to confirm a present value exists.

Across these fields the same move appears: match an unfamiliar series to a familiar one, and let the familiar one settle the question.

What Are The Most Common Mistakes With The Limit Comparison Test?

These three errors account for most lost marks, and each matches a question real students ask on r/calculus and on university convergence-test handouts.

Reading the $L = 0$ or $L = \infty$ case in the wrong direction.

Where it slips in:

A student gets $L = 0$, sees the benchmark $\sum b_n$ diverge, and wrongly concludes the original series diverges too.

Don't do this:

Do not use the $L = 0$ case with a divergent benchmark or the $L = \infty$ case with a convergent one. In both pairings the test is inconclusive and proves nothing.

The correct way:

Match the edge case to its benchmark. Use $L = 0$ only with a convergent $\sum b_n$ to conclude convergence, and $L = \infty$ only with a divergent $\sum b_n$ to conclude divergence. Otherwise pick a new $b_n$.

Applying the test to a series with negative terms.

Where it slips in:

A student runs the Limit Comparison Test on an alternating or sign-changing series such as $\sum \frac{(-1)^n}{n}$, treating it like a positive-term series.

Don't do this:

Do not use this test when the terms are not eventually positive. The theorem's hypotheses require $a_n > 0$ and $b_n > 0$, and the argument breaks without them.

The correct way:

Check the sign first. For alternating terms, use the alternating series test, or apply the Limit Comparison Test to the series of absolute values $\sum |a_n|$ to test for absolute convergence.

Comparing to a benchmark whose fate you do not know.

Where it slips in:

A student picks a $b_n$ that looks convenient, computes a finite positive $L$, and then cannot say anything because they never established whether $\sum b_n$ itself converges.

Don't do this:

Do not compare against a series you have not already classified. A shared fate is useless if the fate of the benchmark is unknown.

The correct way:

Always choose $b_n$ to be a p-series or a geometric series, whose convergence you can read off instantly from $p > 1$ or $|r| < 1$. Keep the dominant powers, and the benchmark that appears is one of these by design.

Practice Problems On The Limit Comparison Test

Decide whether each series converges or diverges using the Limit Comparison Test. Answers are verified.

1. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 4}$.
   (Answer: take $b_n = \frac{1}{n^2}$, $L = 1$; $\sum \frac{1}{n^2}$ converges, so the series converges.)

2. $\displaystyle\sum_{n=1}^{\infty} \frac{n + 1}{n^2 + 2n}$.
   (Answer: behaves like $\frac{n}{n^2} = \frac{1}{n}$, take $b_n = \frac{1}{n}$, $L = 1$; harmonic series diverges, so the series diverges.)

3. $\displaystyle\sum_{n=1}^{\infty} \frac{3n^2 - 1}{n^4 + n}$.
   (Answer: behaves like $\frac{3}{n^2}$, take $b_n = \frac{1}{n^2}$, $L = 3$; converges.)

4. $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 2}}$.
   (Answer: take $b_n = \frac{1}{n^{3/2}}$, $L = 1$; p-series with $p = \frac{3}{2} > 1$ converges, so the series converges.)

5. $\displaystyle\sum_{n=1}^{\infty} \frac{2n}{\sqrt{n^4 + 1}}$.
   (Answer: behaves like $\frac{2n}{n^2} = \frac{2}{n}$, take $b_n = \frac{1}{n}$, $L = 2$; diverges.)

6. $\displaystyle\sum_{n=1}^{\infty} \frac{5}{2^n + n}$.
   (Answer: behaves like $\frac{5}{2^n}$, take $b_n = \left(\frac{1}{2}\right)^n$, $L = 5$; geometric with $|r| < 1$ converges, so the series converges.)

Where Should You Go Next After The Limit Comparison Test?

Once the pick-a-benchmark move feels automatic, the natural doors lead to the neighbouring tests and to the ideas the comparison rests on.

1. The comparison test. The direct inequality version, worth mastering so you know when a clean bound beats a limit.

2. The ratio test. The workhorse for series with powers or an $n$ factorial, where dominant-power comparison runs out of road.

3. The harmonic series. The single most useful divergent benchmark, the one your $b_n = \frac{1}{n}$ keeps landing on.

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