Computing Successive-Term Limits for the Ratio Test

Why Compare Consecutive Terms?

A positive-term series can be difficult to classify by looking at its terms one at a time. The ratio test instead compares each term with the one immediately before it. If the terms eventually shrink by a roughly fixed proportion below one, the series converges; if they grow by a proportion above one, they cannot even approach zero.

The key calculation is the limit of \(a_{n+1}/a_n\). Applying the test correctly requires care with the index shift, cancellation, powers, and the final interpretation. A ratio limit is not the sum of the series: it is information about how its terms behave.

The Ratio Test for Positive Terms

Definition: Let \(\sum_{n=1}^{\infty}a_n\) be a series with \(a_n>0\) for every \(n\). If the limit \(L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\) exists, with \(L\) finite or \(L=+\infty\), then the ratio test gives these conclusions:

If \(0\le L<1\), the series converges.
If \(1<L\le+\infty\), the series diverges.
If \(L=1\), the test gives no conclusion.

To apply the test, write \(a_{n+1}\) by replacing every \(n\) in the formula for \(a_n\) with \(n+1\). Then form the quotient \(a_{n+1}/a_n\), simplify it, and only then evaluate its limit. The order matters: the test compares the next term to the current term, not the current term to the next one.

The test has a strict boundary at one. A limit below one indicates eventual shrinkage at a rate sufficient for convergence. A limit above one, including an infinite limit, means the terms fail to approach zero. At exactly one, the ratio does not distinguish convergence from divergence.

A Useful Simplification for Polynomial Factors

Many calculations contain a polynomial in \(n\) multiplied by a fixed exponential factor. Suppose \(a_n=P(n)c^n\), where \(P\) is a polynomial with positive values for the indices considered and \(c>0\). Direct substitution gives

\frac{a_{n+1}}{a_n} = \frac{P(n+1)c^{n+1}}{P(n)c^n} = c\frac{P(n+1)}{P(n)}.

If \(P\) has degree \(d\) and leading coefficient \(b\ne0\), then \(P(n)/n^d\to b\). Since \(P(n+1)/(n+1)^d\to b\) as well, and \((n+1)^d/n^d\to1\), it follows that \(P(n+1)/P(n)\to1\). Therefore the ratio limit in this setting is \(c\).

This short derivation explains a common cancellation pattern: shifting a polynomial factor by one usually does not change its leading large-\(n\) scale, while the fixed exponential factor contributes the lasting multiplier. It is still necessary to write and simplify the actual ratio; the pattern does not replace the calculation when the term has a different form.

Worked Examples

Worked Example: A Polynomial Factor with an Exponential Denominator

Determine whether the positive-term series with \(a_n=\frac{n^2+1}{4^n}\) converges.

First shift the index in the entire term formula:

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

Now divide this next term by the current term. Dividing by a fraction means multiplying by its reciprocal:

\frac{a_{n+1}}{a_n} = \frac{n^2+2n+2}{4^{n+1}} \cdot \frac{4^n}{n^2+1} = \frac{n^2+2n+2}{4(n^2+1)}.

Divide numerator and denominator by \(n^2\) to evaluate the limit:

\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{1+\frac{2}{n}+\frac{2}{n^2}} {4\left(1+\frac{1}{n^2}\right)} = \frac14.

Since \(1/4<1\), the ratio test says that the series converges.

Check: Substituting \(n+1\) into the original expression gives the numerator \((n+1)^2+1=n^2+2n+2\) and the denominator \(4^{n+1}=4\cdot4^n\). Dividing by \(a_n\) cancels \(4^n\), leaving exactly the factor \(1/4\) and the displayed polynomial quotient.

Worked Example: A Limit Greater Than One

Determine whether the series with \(a_n=\frac{3^n}{n+1}\) converges. Each term is positive for \(n\ge1\).

Shift the index in both the exponential and the denominator:

a_{n+1} = \frac{3^{n+1}}{n+2}.

Form and simplify the successive-term ratio:

\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{n+2}\cdot\frac{n+1}{3^n} = 3\frac{n+1}{n+2}.

Dividing the fraction by \(n\) in its numerator and denominator shows its limit:

\lim_{n\to\infty}3\frac{n+1}{n+2} = 3\lim_{n\to\infty}\frac{1+\frac1n}{1+\frac2n} = 3.

Because \(3>1\), the ratio test says that the series diverges.

Check: The quotient of the exponential factors is \(3^{n+1}/3^n=3\), and the quotient of the remaining factors is \((n+1)/(n+2)\). Their product is the ratio above, which tends to \(3\).

Worked Example: An Infinite Ratio Limit

Consider the positive terms \(a_n=2^{n^2}\). Determine the ratio limit and the series conclusion.

Replace \(n\) by \(n+1\), then divide by the original term:

a_{n+1}=2^{(n+1)^2}, \qquad \frac{a_{n+1}}{a_n} = \frac{2^{(n+1)^2}}{2^{n^2}} = 2^{(n+1)^2-n^2} = 2^{2n+1}.

As \(n\) increases, the exponent \(2n+1\) increases without bound, so \(2^{2n+1}\to+\infty\). Thus \(L=+\infty\), and the ratio test says that the series diverges.

Check: Expanding \((n+1)^2\) gives \(n^2+2n+1\); subtracting the denominator exponent \(n^2\) leaves \(2n+1\), as in the simplified ratio. Also, \(a_n=2^{n^2}\) itself does not tend to zero, which is consistent with divergence.

Worked Example: A Ratio Limit Equal to One

The ratio test’s boundary case is important because \(L=1\) does not settle convergence. Compare \(a_n=1/n\) with \(b_n=1/n^2\). Both sequences have positive terms for \(n\ge1\).

For the first series,

\frac{a_{n+1}}{a_n} = \frac{1}{n+1}\cdot n = \frac{n}{n+1}, \qquad \lim_{n\to\infty}\frac{n}{n+1}=1.

This is the harmonic series, which diverges. For the second series,

\frac{b_{n+1}}{b_n} = \frac{1}{(n+1)^2}\cdot n^2 = \left(\frac{n}{n+1}\right)^2, \qquad \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^2=1.

This is a \(p\)-series with \(p=2\), so it converges. These standard examples have the same ratio limit but different convergence behavior. The ratio test is inconclusive for both.

Check: In the first quotient, dividing \(1/(n+1)\) by \(1/n\) gives \(n/(n+1)\). In the second, dividing \(1/(n+1)^2\) by \(1/n^2\) gives \(n^2/(n+1)^2\). Each limit is one, so neither conclusion can be obtained from the ratio test.

A Reliable Calculation Process

For a new positive-term series, the following sequence of steps reduces errors and keeps the calculation transparent.

Identify the term.
Write the general term as \(a_n\), and check that \(a_n>0\) for the indices in question.

Shift the index.
Write \(a_{n+1}\) by replacing each occurrence of \(n\) in the term formula with \(n+1\).

Form the ratio in the correct order.
Calculate \(a_{n+1}/a_n\), keeping parentheses and exponents intact.

Simplify before taking the limit.
Cancel common factors and rewrite powers using exponent rules. Do not discard terms until their limiting contribution is clear.

Interpret the limit.
Compare the result with one: below one means convergence, above one or infinity means divergence, and exactly one is inconclusive.

Common Mistake: Reversing the Ratio

A frequent error is to compute \(a_n/a_{n+1}\) instead of the required \(a_{n+1}/a_n\). In the first worked example, the correct ratio is

\frac{a_{n+1}}{a_n} = \frac{n^2+2n+2}{4(n^2+1)} \longrightarrow \frac14.

If the quotient is reversed, its limit is \(4\), since it is the reciprocal of the correct ratio. Applying the ratio test to that reversed quotient would incorrectly suggest divergence. The definition specifically uses the next term divided by the current term. When in doubt, write both \(a_{n+1}\) and \(a_n\) explicitly before making the division.

Why the Limit Is Useful—and Where It Stops

The ratio test compresses information about an entire sequence of terms into one limiting comparison. For instance, a limit of \(1/4\) tells us that, for sufficiently large \(n\), consecutive terms shrink at a proportion close to \(1/4\). A limit of \(3\) means the terms eventually increase by a proportion close to \(3\). This is why the distinction around one matters.

However, the limit is not a general verdict when it equals one. The harmonic series and the convergent \(p\)-series in the boundary example demonstrate that terms can have successive-term ratios approaching one while the series behave differently. In that case, another test or a different argument is needed. The correct conclusion is “inconclusive,” not “convergent” or “divergent.”

Finally, divergence from the ratio test is consistent with a necessary condition for convergence: the terms of a convergent series must tend to zero. A ratio limit greater than one or infinite forces the terms not to tend to zero. A ratio limit below one gives the stronger conclusion of convergence. Keeping these roles distinct prevents a ratio calculation from being mistaken for a calculation of the series’ sum.

Check Your Understanding

For each question, state the ratio limit and the conclusion the ratio test permits.

For \(a_n=\frac{n+3}{5^n}\), write and simplify \(a_{n+1}/a_n\). What is its limit?
If a positive-term series has ratio limit \(L=0.8\), what does the ratio test conclude?
If the ratio limit is \(L=1.2\), what does the ratio test conclude, and what necessary condition for convergence is relevant?
For \(a_n=1/n^3\), calculate \(a_{n+1}/a_n\) and its limit. Can the ratio test decide convergence?
What is wrong with using \(a_n/a_{n+1}\) in place of \(a_{n+1}/a_n\)?