The Comparison Test

When Comparing Ratios Is Easier

The previous tutorial showed how an inequality between positive terms controls the corresponding partial sums. Sometimes the needed inequality is easy to find directly. In other problems, however, the terms have complicated expressions, and it is not obvious how to bound one by another for every index. A useful alternative is to compare their ratio.

If two positive terms have a ratio that approaches a finite, strictly positive number, then, for all sufficiently large indices, each term is bounded above and below by a positive constant times the other. The terms need not be close to equal. What matters is that neither series eventually becomes arbitrarily small relative to the other. This observation leads to the limit comparison test.

As before, a positive-term series means a series whose terms are nonnegative. For ratio comparisons, we will assume the terms being divided are strictly positive for all sufficiently large indices. A finite number of initial terms does not determine convergence: the earlier Theorem (Finite Changes Preserve Series Convergence) lets us focus on the tail where the ratio is defined and the relevant inequalities hold.

The Limit Comparison Test

Theorem (Limit Comparison Test): Suppose $a_n>0$ and $b_n>0$ for all sufficiently large $n$, and suppose $$ \lim_{n\to\infty}\frac{a_n}{b_n}=L $$ for some $L$ with $0<L<\infty$. Then $\sum_{n=1}^{\infty}a_n$ converges if and only if $\sum_{n=1}^{\infty}b_n$ converges. Equivalently, either both series converge or both diverge.

Proof. Since $a_n/b_n$ tends to $L>0$, take $\varepsilon=L/2$ in the definition of convergence of a sequence. There is an integer $N$ such that, for every $n\geq N$,

\left|\frac{a_n}{b_n}-L\right|<\frac{L}{2}.

It follows that $L/2<a_n/b_n<3L/2$. Since $b_n>0$, multiplication by $b_n$ preserves the inequalities:

\frac{L}{2}b_n<a_n<\frac{3L}{2}b_n \qquad(n\geq N).

If $\sum b_n$ converges, then the series with terms $(3L/2)b_n$ converges by the Linearity of Convergent Series. On the tail, $0\leq a_n\leq(3L/2)b_n$, so the Comparison Principle for Positive-Term Series shows that the tail of $\sum a_n$ converges. Adding or removing the finitely many terms before $N$ does not affect convergence, by the Theorem (Finite Changes Preserve Series Convergence). Thus $\sum a_n$ converges.

Conversely, if $\sum a_n$ converges, then the lower bound gives $0\leq b_n<(2/L)a_n$ for $n\geq N$. The series with terms $(2/L)a_n$ converges by linearity, so the Comparison Principle shows that the tail of $\sum b_n$ converges. The finite initial terms again do not affect convergence. Thus $\sum b_n$ converges. We have proved both implications, so the two series have the same convergence behavior. $\square$

The proof explains why the limit must be strictly between zero and infinity. A finite positive limit provides two eventual bounds, one in each direction. Those bounds let us transfer convergence either way. If the ratio tends to zero or infinity, one of the two bounds is lost, and only a one-directional conclusion may be available.

Worked Example: Comparing a Rational Term with a Square Reciprocal

Consider the series whose $n$th term is $(3n+2)/(n^3+4n)$. Compare it with $b_n=1/n^2$. Both terms are positive for $n\geq1$, and their ratio is

\frac{(3n+2)/(n^3+4n)}{1/n^2} =\frac{3n^2+2n}{n^2+4}.

Dividing the numerator and denominator on the right by $n^2$ shows that the ratio tends to $3$. The Limit Comparison Test therefore reduces the question to the convergence of $\sum 1/n^2$. For completeness, when $n\geq2$,

0<\frac{1}{n^2}\leq\frac{1}{n(n-1)} =\frac{1}{n-1}-\frac{1}{n}.

The series on the right telescopes: its partial sum from $n=2$ through $n=N$ is $1-1/N$, which is bounded above. Its terms are nonnegative, so the earlier Theorem (Bounded Increasing Partial Sums) shows that this comparison series converges. The Comparison Principle then gives convergence of $\sum 1/n^2$, and the Limit Comparison Test gives convergence of $\sum_{n=1}^{\infty}(3n+2)/(n^3+4n)$.

Worked Example: A Rational Term with Harmonic Behavior

Now consider the terms $(5n^2+1)/(2n^3+7n)$. Compare them with the harmonic terms $b_n=1/n$. The ratio simplifies to

\frac{(5n^2+1)/(2n^3+7n)}{1/n} =\frac{5n^2+1}{2n^2+7}.

Dividing the numerator and denominator by $n^2$ gives a limit of $5/2$, which is finite and strictly positive. The Limit Comparison Test says that the two series have the same behavior. The harmonic series $\sum 1/n$ diverges, as shown by grouping its terms into blocks in the previous tutorial. Therefore

\sum_{n=1}^{\infty}\frac{5n^2+1}{2n^3+7n} \quad\text{diverges.}

The ratio need not tend to $1$. A limit of $5/2$ is enough: it means the terms are eventually within fixed positive constant factors of one another.

Ratios Tending to Zero or Infinity

The finite-positive-limit test has useful one-sided versions. If $a_n/b_n\to0$, then $a_n$ is eventually no larger than $b_n$. This can transfer convergence from $b_n$ to $a_n$, but it cannot generally transfer divergence from $b_n$ to $a_n$. If $a_n/b_n\to\infty$, then $a_n$ is eventually no smaller than $b_n$. This can transfer divergence from $b_n$ to $a_n$, but it cannot generally transfer convergence from $a_n$ to $b_n$.

Theorem (One-Sided Limit Comparison): Suppose $a_n>0$ and $b_n>0$ for all sufficiently large $n$.

If $a_n/b_n\to0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
If $a_n/b_n\to\infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

Proof. In the first case, the definition of $a_n/b_n\to0$, with $\varepsilon=1$, gives an index $N$ such that $0<a_n/b_n<1$ for $n\geq N$. Multiplication by $b_n>0$ gives $0<a_n<b_n$. The Comparison Principle shows that the tail of $\sum a_n$ converges, and finite changes preserve convergence.

In the second case, the definition of $a_n/b_n\to\infty$ gives an index $N$ such that $a_n/b_n>1$ for every $n\geq N$. Hence $a_n>b_n$ on that tail. The tail of $\sum b_n$ diverges, since removing finitely many terms cannot turn a divergent positive-term series into a convergent one. The divergence part of the Comparison Principle now shows that the tail of $\sum a_n$ diverges. Therefore $\sum a_n$ diverges as well. $\square$

Worked Example: A Ratio Tending to Zero

Consider $a_n=1/(n2^n)$ and compare it with $b_n=1/2^n$. Their ratio is $a_n/b_n=1/n$, which tends to zero. The geometric series $\sum 1/2^n$ converges because its ratio is $1/2$, whose absolute value is less than $1$. The first conclusion of the One-Sided Limit Comparison Theorem therefore gives

\sum_{n=1}^{\infty}\frac{1}{n2^n} \quad\text{converges.}

This argument uses the convergence of the larger comparison terms. A ratio tending to zero alone does not decide convergence: it says that $a_n$ is eventually smaller than $b_n$, not that either series has a particular behavior.

Worked Example: A Ratio Tending to Infinity

Let $a_n=1/\sqrt{n}$ and $b_n=1/n$. For every $n\geq1$, both terms are positive, and

\frac{a_n}{b_n} =\frac{1/\sqrt{n}}{1/n} =\sqrt{n}\longrightarrow\infty.

The harmonic series $\sum 1/n$ diverges. The second conclusion of the One-Sided Limit Comparison Theorem therefore shows that $\sum 1/\sqrt{n}$ diverges. Indeed, the ratio argument says that $1/\sqrt{n}\geq1/n$ for every $n\geq1$, so the divergent harmonic series supplies a lower bound.

Choosing and Interpreting a Comparison

A useful comparison series should have known behavior and a ratio that is manageable. For rational expressions, the highest powers often reveal the likely scale: after forming $a_n/b_n$, divide numerator and denominator by the highest relevant power of $n$. For products involving exponentials, factorials, or other familiar terms, simplify the ratio algebraically before deciding which test applies.

Keep the direction of implication in view. A ratio tending to zero says the $a_n$ terms are smaller in the long run. If the $b_n$ series converges, that is enough to prove convergence of the smaller series. But if the $b_n$ series diverges, the smaller series might still converge. For example, $1/2^n\leq1/n$ for every $n\geq1$, while the geometric series converges and the harmonic series diverges. Thus a divergent upper comparison does not settle the behavior of the lower series.

Likewise, when the ratio tends to infinity, the larger terms inherit divergence from a divergent lower comparison, but convergence of the larger series is not established by convergence of the smaller one. Do not treat either one-sided conclusion as an equivalence. An equivalence is justified by the Limit Comparison Test only when the ratio has a finite, strictly positive limit.

Choose a familiar comparison.
Select positive terms whose series behavior is already known, such as geometric or harmonic terms.

Form the ratio.
Simplify $a_n/b_n$, checking that the terms are positive wherever division is used.

Evaluate its limit.
A finite positive limit gives equivalent behavior; a zero or infinite limit gives only the corresponding one-sided conclusion.

Apply the correct direction.
Use a convergent upper comparison to prove convergence, or a divergent lower comparison to prove divergence. Ignore only finitely many initial terms when justified by finite changes.

Takeaway: If the ratio of two positive terms tends to a finite positive number, their series either both converge or both diverge. A ratio tending to zero or infinity can still be useful, but it supports only a one-sided conclusion and must be paired with a comparison series of the appropriate behavior.

Check Your Understanding

Use the ratio limit and the direction of comparison to answer these questions.

If $a_n/b_n\to L$ with $0<L<\infty$, what relationship holds between the convergence behavior of the two series?
Why does a finite positive ratio limit provide bounds in both directions for all sufficiently large $n$?
If $a_n/b_n\to0$ and $\sum b_n$ converges, what conclusion follows? Does the same ratio limit settle the case when $\sum b_n$ diverges?
If $a_n/b_n\to\infty$ and $\sum b_n$ diverges, which comparison principle gives the conclusion?
Why can finitely many terms be excluded while applying a limit comparison argument?

What You'll Learn