Limit Comparison Test

Why Compare Ratios?

The Comparison Test works directly with inequalities, but finding a useful inequality can be awkward. The Limit Comparison Test often replaces that task with a limit calculation: instead of proving that one term is smaller than another, we determine whether their ratio approaches a positive finite number. If it does, the two positive series have the same convergence behavior.

The earlier Theorem (Limit Comparison Test) states that if $a_n$ and $b_n$ are positive for all sufficiently large $n$, and

\lim_{n\to\infty}\frac{a_n}{b_n}=L,\qquad 0<L<\infty,

then $\sum a_n$ converges if and only if $\sum b_n$ converges. The one-sided form established earlier also says that if $a_n/b_n\to0$ and $\sum b_n$ converges, then $\sum a_n$ converges. Here we focus on how to select $b_n$, compute the relevant limits efficiently, and understand which conclusions the limit does—and does not—justify.

Reading the Ratio and Choosing a Benchmark

A useful comparison series has terms that capture the dominant size of $a_n$. Familiar benchmarks include geometric terms $r^n$, harmonic terms $1/n$, and terms such as $1/n^2$. The goal is not to find a benchmark equal to $a_n$; it is to make the ratio simple and give the benchmark a known convergence behavior.

A positive finite ratio limit is the strongest situation. If $a_n/b_n\to L$ with $L>0$, then eventually the ratio is bounded above and below by positive constants. For example, eventually $L/2<a_n/b_n<2L$. This explains why the two series have the same behavior: their terms differ by at most fixed positive factors in the tail. The formal transfer of convergence is the earlier Limit Comparison Test.

A ratio tending to zero is different. It says that $a_n$ is small relative to $b_n$, so the one-sided test can transfer convergence from $b_n$ to $a_n$. It does not imply that $\sum a_n$ converges when $\sum b_n$ diverges. For instance, $1/2^n$ divided by $1/n$ tends to zero, but that fact alone does not transfer a conclusion from the divergent harmonic series; in this particular example, convergence is instead known from the geometric-series theorem.

When a ratio tends to infinity, reverse the roles if appropriate: $a_n/b_n\to+\infty$ means $b_n/a_n\to0$. This can be useful if $\sum a_n$ is known to converge and the one-sided test applies to $b_n$. But an infinite ratio is not itself a finite positive limit, so the two-sided Limit Comparison Test cannot be applied to it directly.

Leading Terms in Polynomial Ratios

For expressions involving polynomials, the highest powers determine the limiting ratio. The following lemma makes that calculation precise and is useful well beyond series tests.

Lemma (Leading-Term Quotient Limit): Let $P$ and $Q$ be nonzero real polynomials of degrees $d$ and $e$, with leading coefficients $p$ and $q$, respectively. Then, for positive integers $n$ sufficiently large that $Q(n)\ne0$, $$ \lim_{n\to\infty}\frac{P(n)}{Q(n)n^{d-e}}=\frac{p}{q}. $$

Proof. Write the polynomials in descending powers:

P(n)=p n^d+p_{d-1}n^{d-1}+\cdots+p_0,\qquad Q(n)=q n^e+q_{e-1}n^{e-1}+\cdots+q_0.

Dividing the first expression by $n^d$ and the second by $n^e$ gives

\frac{P(n)}{n^d}=p+\frac{p_{d-1}}{n}+\cdots+\frac{p_0}{n^d}\longrightarrow p, \qquad \frac{Q(n)}{n^e}=q+\frac{q_{e-1}}{n}+\cdots+\frac{q_0}{n^e}\longrightarrow q.

Since $q\ne0$, the denominator $Q(n)/n^e$ is nonzero for all sufficiently large $n$, and the quotient limit law yields

\frac{P(n)}{Q(n)n^{d-e}} = \frac{P(n)/n^d}{Q(n)/n^e} \longrightarrow\frac{p}{q}.

This proves the lemma. $\square$

In particular, if $d<e$, then $P(n)/Q(n)$ behaves like a constant multiple of $1/n^{e-d}$. If the difference in degrees is one, the natural benchmark is the harmonic term $1/n$. If the difference is at least two, a natural benchmark is $1/n^2$ or a smaller positive term. If $d\geq e$, the quotient does not tend to zero, so the necessary condition for series convergence can often settle the question immediately.

A Degree Test for Positive Rational Terms

Theorem (Degree Classification for Positive Rational Terms): Let $P$ and $Q$ be nonzero real polynomials of degrees $d$ and $e$, with positive leading coefficients. Suppose $a_n=P(n)/Q(n)$ is positive for all sufficiently large $n$. Then:

If $e-d=1$, the series $\sum a_n$ diverges.
If $e-d\geq2$, the series $\sum a_n$ converges.
If $e-d\leq0$, the series $\sum a_n$ diverges.

Proof. Put $k=e-d$. By the Leading-Term Quotient Limit,

\frac{a_n}{1/n^k} = \frac{P(n)}{Q(n)n^{-k}} \longrightarrow\frac{p}{q},

where $p$ and $q$ are the positive leading coefficients. If $k=1$, the comparison term is $1/n$, and the limit $p/q$ is positive and finite. The Limit Comparison Test and divergence of the harmonic series imply that $\sum a_n$ diverges.

If $k\geq2$, then $1/n^k\leq1/n^2$ for every $n\geq1$. The series $\sum 1/n^2$ converges: for $n\geq2$,

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

and the partial sums of the right-hand side telescope and are bounded. The Comparison Test therefore gives convergence of $\sum 1/n^k$. The positive finite ratio limit between $a_n$ and $1/n^k$, together with the Limit Comparison Test, gives convergence of $\sum a_n$.

Finally, if $k\leq0$, then $d-e=-k\geq0$. The polynomial quotient $P(n)/Q(n)$ tends to $p/q>0$ when $d=e$, and tends to $+\infty$ when $d>e$. In either case $a_n$ does not tend to zero. The Necessary Condition for Series Convergence therefore implies that $\sum a_n$ diverges. These cases establish the theorem. $\square$

The positivity assumption matters. Positive leading coefficients ensure the rational terms are positive for sufficiently large $n$, as required for the positive-series tests. If a rational expression changes sign infinitely often, this degree classification does not by itself decide convergence of the signed series.

Worked Applications

Worked Example: A Radical Term Compared with the Harmonic Series

Determine the behavior of

\sum_{n=1}^{\infty} \frac{\sqrt{9n^2+2n}+1}{n^2+5n}.

The numerator grows like $3n$ and the denominator like $n^2$, suggesting $1/n$ as a benchmark. Let

a_n=\frac{\sqrt{9n^2+2n}+1}{n^2+5n}, \qquad b_n=\frac1n.

Both terms are positive. Dividing numerator and denominator in the ratio by the appropriate powers of $n$ gives

\frac{a_n}{b_n} = \frac{n\bigl(\sqrt{9n^2+2n}+1\bigr)}{n^2+5n} = \frac{\sqrt{9+2/n}+1/n}{1+5/n} \longrightarrow 3.

The limit is positive and finite. Since $\sum 1/n$ diverges, the Limit Comparison Test proves that the displayed series diverges.

Worked Example: A Rational Series with a Degree Difference of Two

Consider

\sum_{n=1}^{\infty}\frac{7n^2+3n+1}{2n^4+n^2+5}.

The degree difference is $4-2=2$, so compare with $1/n^2$. The ratio is

\frac{\dfrac{7n^2+3n+1}{2n^4+n^2+5}}{1/n^2} = \frac{7n^4+3n^3+n^2}{2n^4+n^2+5} = \frac{7+3/n+1/n^2}{2+1/n^2+5/n^4} \longrightarrow\frac72.

The terms are positive, and the ratio limit is positive and finite. The series $\sum 1/n^2$ converges, so the Limit Comparison Test proves convergence of the given series.

Worked Example: A Zero Ratio with a Geometric Comparison

Determine the behavior of

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

Choose the convergent geometric series with comparison terms $b_n=1/2^n$. For every $n\geq1$, the given term is positive, and

\frac{\dfrac{n+1}{n^2\,2^n}}{1/2^n} = \frac{n+1}{n^2} = \frac1n+\frac1{n^2} \longrightarrow0.

The geometric series $\sum 1/2^n$ converges because its ratio is $1/2$, whose absolute value is less than one. The one-sided Limit Comparison Test now proves that the series in the question converges. Notice that the ratio limit is zero, not positive: this is why we use the one-sided form rather than the two-sided Limit Comparison Test.

Limits That Do Not Decide the Series

A limit calculation is useful only when paired with a test whose hypotheses fit that limit. In particular, a ratio tending to zero does not always settle convergence. For example, take $a_n=1/n^2$ and $b_n=1/n$. Then $a_n/b_n=1/n\to0$, while the benchmark series $\sum b_n$ diverges. The one-sided test has no conclusion in this direction, even though $\sum a_n$ happens to converge.

Likewise, if $a_n/b_n\to+\infty$, do not apply the finite-positive-limit version as if infinity were an allowed value. Instead, consider the reciprocal ratio and ask whether the one-sided test can be applied with the roles exchanged. If neither the reciprocal comparison nor another test gives a valid conclusion, the ratio calculation alone is inconclusive.

A separate issue is sign. The Limit Comparison Test for positive series requires positivity eventually. A finite number of nonpositive initial terms does not affect convergence, by the earlier Theorem (Finite Changes Preserve Series Convergence), but the terms must be positive from some point onward for the positive-series argument to apply. Always check this before taking a ratio or transferring a conclusion.

Check eventual positivity.
Confirm that the terms and the proposed comparison terms are positive for all sufficiently large indices.

Identify the dominant size.
For polynomials, retain the highest powers; for a geometric factor, consider a geometric benchmark.

Compute the term ratio.
Simplify by factoring out dominant powers, and evaluate the limit carefully.

Match the limit to the right result.
Use the two-sided test for a positive finite limit and the one-sided test for a zero limit with a convergent comparison.

Takeaway: The Limit Comparison Test is most effective when the benchmark captures the dominant size of the terms. A positive finite ratio limit transfers convergence behavior in both directions; a zero limit transfers convergence only from a convergent comparison series to the smaller terms.

Check Your Understanding

For each question, identify what the ratio limit permits you to conclude and which hypotheses must be checked.

If $a_n,b_n>0$ eventually and $a_n/b_n\to4$, what relationship holds between the convergence behavior of the two series?
If $a_n/b_n\to0$ and $\sum b_n$ converges, what conclusion follows? Does the same conclusion follow if $\sum b_n$ diverges?
For a positive rational term $P(n)/Q(n)$ with degree difference $\deg Q-\deg P=3$, which standard series is a natural comparison?
Why does a positive rational term that does not tend to zero give a divergent series?
What must be done before using the Limit Comparison Test if the ratio appears to tend to infinity?

What You'll Learn