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Ratio, Root, and Test Selection · Tutorial 693 of 998

Using Ratio and Root Tests for Absolute Convergence

Use the ratio or root test on the magnitudes of signed terms, then justify why a strict test result gives absolute convergence of the original series.

Advanced 8 min read

What You'll Learn

  • Define absolute convergence for a signed series.
  • Apply the ratio test to the magnitudes of its terms.
  • Apply the root test when the terms have convenient nth roots.
  • Use a strict limit below one to prove absolute convergence.
  • Recognize why a signed ratio or a boundary limit is not enough.

Why Test the Magnitudes of a Signed Series?

A signed series can have positive and negative terms, so its partial sums may reflect both the sizes of the terms and their signs. To determine whether the series converges absolutely, we set the signs aside and examine the series of magnitudes. The ratio and root tests are especially useful when those magnitudes contain factorials, powers, or expressions raised to the index.

The key distinction is that a test applied to the magnitudes answers a specific question: does the series of absolute values converge? When it does, the original signed series converges as well. The conclusion is stronger than merely establishing convergence of the signed series.

Definition: A series \(\sum_{n=1}^{\infty} a_n\) is absolutely convergent if the series of absolute values \(\sum_{n=1}^{\infty}|a_n|\) converges. It is enough to check the magnitudes \(|a_n|\); the signs of the original terms do not affect this definition.

The Strict Ratio and Root Conditions

Write \(b_n=|a_n|\), so \(b_n\ge 0\). The ratio test and root test are applied to this nonnegative sequence. Their strict convergence conditions are as follows.

Theorem: Suppose \(b_n=|a_n|\). If \(b_n>0\) for all sufficiently large \(n\) and \(\displaystyle \lim_{n\to\infty}\frac{b_{n+1}}{b_n}=L<1\), then \(\sum b_n\) converges, and therefore \(\sum a_n\) converges absolutely. If \(\displaystyle \lim_{n\to\infty}\sqrt[n]{b_n}=L<1\), then the same conclusion holds. In either test, the strict inequality \(L<1\) is essential to this convergence conclusion.

In the ratio test, the condition that \(b_n\) is eventually positive makes the quotient well-defined. If terms are eventually zero, then the absolute-value series has only finitely many nonzero terms and converges directly. When the root limit exists, the root test does not require division by \(b_n\).

The two tests are often summarized by comparing a limiting size with one. A limit strictly below one gives absolute convergence. A limit equal to one does not settle the question using that test. In particular, do not replace the strict condition with a non-strict one.

Why a Strict Limit Below One Works

Both tests work by bounding the absolute values by the terms of a convergent geometric series. Here is the argument for each test.

For the ratio test, suppose \(b_{n+1}/b_n\to L<1\). Choose a number \(q\) with \(L<q<1\). By the definition of a limit, there is an index \(N\) such that for every \(n\ge N\), \(b_{n+1}/b_n<q\). Repeatedly applying this inequality gives

$$ b_{N+1}<b_Nq,\qquad b_{N+2}<b_Nq^2,\qquad \ldots,\qquad b_{N+k}<b_Nq^k. $$

The tail of \(\sum b_n\) is therefore bounded by a constant multiple of a geometric series with ratio \(q<1\), so that tail converges. Adding the finitely many terms before \(N\) does not change convergence.

For the root test, suppose \(\sqrt[n]{b_n}\to L<1\). Choose \(q\) with \(L<q<1\). For all sufficiently large \(n\), \(\sqrt[n]{b_n}<q\), and hence \(b_n<q^n\). The tail is bounded term by term by a convergent geometric series. This proves convergence of \(\sum b_n\).

Finally, convergence of \(\sum |a_n|\) implies convergence of \(\sum a_n\). Indeed, for any two partial sums with indices \(p>q\), the triangle inequality gives

$$ \left|\sum_{n=q+1}^{p}a_n\right| \le \sum_{n=q+1}^{p}|a_n|. $$

Since the tails of the convergent series \(\sum |a_n|\) can be made arbitrarily small, the tails of \(\sum a_n\) can also be made arbitrarily small. Thus the partial sums of \(\sum a_n\) satisfy the Cauchy criterion and converge. This is why a strict ratio or root result for the magnitudes proves absolute convergence of the original series.

Worked Examples

Worked Example: A Signed Factorial Series with the Ratio Test

Determine whether the series

$$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{5^n}{n!} $$

converges absolutely.

First take the absolute value of the general term. Since \(5^n/n!\) is positive, set \(b_n=5^n/n!\). Compute the ratio using the next term:

$$ \frac{b_{n+1}}{b_n} = \frac{5^{n+1}}{(n+1)!}\cdot\frac{n!}{5^n} = \frac{5}{n+1}. $$

As \(n\to\infty\), \(5/(n+1)\to 0\). The ratio limit is \(L=0<1\), so the ratio test shows that \(\sum b_n\) converges. Since \(b_n\) is exactly the absolute value of the original term, the original series converges absolutely. The check is that the test was applied to \(b_n=|a_n|\), not to the alternating sign factor.

Worked Example: A Variable Base with the Root Test

Determine whether the series

$$ \sum_{n=1}^{\infty}(-1)^n \left(\frac{n+2}{4n+1}\right)^n $$

converges absolutely.

The fraction \((n+2)/(4n+1)\) is positive for \(n\ge1\), so the absolute value of the \(n\)th term is

$$ b_n=\left(\frac{n+2}{4n+1}\right)^n. $$

Take its \(n\)th root:

$$ \sqrt[n]{b_n} = \sqrt[n]{\left(\frac{n+2}{4n+1}\right)^n} = \frac{n+2}{4n+1}. $$

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

$$ \lim_{n\to\infty}\frac{n+2}{4n+1} = \lim_{n\to\infty}\frac{1+2/n}{4+1/n} = \frac14. $$

The root limit is \(1/4<1\). Therefore \(\sum b_n\) converges, and the original signed series converges absolutely. The calculation checks out because taking the \(n\)th root cancels the outer exponent exactly, leaving a quantity with a limit below one.

Worked Example: A Non-Alternating Sign Pattern and the Root Test

Let the signs vary according to \((-1)^{n(n+1)/2}\). Determine whether

$$ \sum_{n=1}^{\infty} (-1)^{n(n+1)/2}\frac{(1+1/n)^n}{5^n} $$

converges absolutely.

The sign factor has absolute value one. Thus the absolute value of the \(n\)th term is

$$ b_n=\frac{(1+1/n)^n}{5^n}. $$

Take the \(n\)th root of the whole expression, keeping track of both factors:

$$ \sqrt[n]{b_n} = \frac{\sqrt[n]{(1+1/n)^n}}{\sqrt[n]{5^n}} = \frac{1+1/n}{5}. $$

Now take the limit:

$$ \lim_{n\to\infty}\frac{1+1/n}{5} = \frac15. $$

Because \(1/5<1\), the root test proves that the absolute-value series converges. Hence the original series converges absolutely. This conclusion does not depend on the particular pattern of signs: the signs disappear when taking absolute values, and the root limit is strictly below one.

A Common Mistake: Using the Signed Ratio

The ratio test for absolute convergence uses the ratio of magnitudes, not the signed quotient \(a_{n+1}/a_n\). For instance, consider the alternating harmonic series \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\). Its signed term ratio is

$$ \frac{a_{n+1}}{a_n} = -\frac{n}{n+1} \longrightarrow -1. $$

A mistaken argument might say that \(-1<1\), so the series converges absolutely. That is not a valid use of the ratio test: the test compares the magnitude of the ratio with one. Here the correct calculation is

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{n}{n+1} \longrightarrow 1. $$

This limit is at the boundary, so the ratio test does not establish absolute convergence. The correction is to take absolute values before forming the ratio and then interpret only the resulting strict condition. A limit of one is inconclusive; it is not evidence either for or against convergence.

Choosing and Applying the Test Carefully

For a signed series, begin by writing the magnitude \(b_n=|a_n|\). Then inspect the form of \(b_n\). A quotient \(b_{n+1}/b_n\) is often effective when factorials or products simplify between consecutive terms. The \(n\)th root is often effective when the term is a quantity raised to the \(n\)th power. In either case, evaluate the limit only after simplifying the expression correctly.

1
Remove the signs.
Write \(b_n=|a_n|\). Absolute convergence concerns \(\sum b_n\).
2
Choose a useful expression.
Use \(b_{n+1}/b_n\) when consecutive terms simplify, or \(\sqrt[n]{b_n}\) when the term has an \(n\)th-power structure.
3
Check the strict condition.
A limit \(L<1\) proves convergence of \(\sum b_n\). A limit equal to one gives no conclusion from that test.
4
State the conclusion precisely.
Convergence of \(\sum |a_n|\) means that the original series \(\sum a_n\) converges absolutely.

Keeping these steps distinct prevents two common errors: applying a test to signed terms instead of magnitudes, and treating a boundary limit as a convergence result. The purpose of the strict tests is not to analyze signs; it is to establish convergence of the absolute-value series, which then guarantees convergence of the original series.

Check Your Understanding

For each question, focus on whether the test is being applied to the absolute values and whether its limit is strictly below one.

  1. What series must converge for \(\sum a_n\) to be absolutely convergent?
  2. If \(\lim_{n\to\infty}|a_{n+1}|/|a_n|=0.8\), what conclusion follows about \(\sum a_n\), provided the terms are eventually nonzero?
  3. Why does \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=1\) fail to establish absolute convergence by the root test?
  4. For \(b_n=(3n+1)^n/(8n)^n\), simplify \(\sqrt[n]{b_n}\) before evaluating its limit.
  5. Why is a signed limit of \(a_{n+1}/a_n\) not, by itself, the ratio-test condition for absolute convergence?

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