When the Ratio Test Is Stronger Than the Root Test

Why Compare the Ratio and Root Tests?

The ratio test examines how each term compares with the term immediately before it. The root test examines the size of a term after taking its \(n\)th root. These are different calculations, but their limits are closely related: if positive successive-term ratios approach a finite value, then the \(n\)th roots approach that same value.

This relationship explains why a successful ratio-test calculation also predicts what the root test would conclude. It does not mean that the tests are interchangeable in every situation, or that either test settles a series when the relevant limit is 1. The hypotheses and the boundary case matter.

Central result: If \(a_n>0\) and the successive-term ratios \(a_{n+1}/a_n\) converge to a finite limit \(L\), then \(a_n^{1/n}\) also converges to \(L\).

The Relationship in Precise Terms

Let \((a_n)\) be a sequence of positive real numbers. Its successive-term ratio at index \(n\) is \(a_{n+1}/a_n\), and its \(n\)th root is \(a_n^{1/n}\). Since all the ratios are positive, any finite limit \(L\) of those ratios must satisfy \(L\geq 0\).

Theorem: Suppose \(a_n>0\) for every \(n\), and \[ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L \] for a finite real number \(L\). Then \[ \lim_{n\to\infty}a_n^{1/n}=L. \]

The idea is that a term far along the sequence is obtained by multiplying together many successive ratios. If each of those ratios is close to \(L\), their product is bounded between products of numbers close to \(L\). Taking \(n\)th roots turns those products into bounds close to \(L\) itself. The proof makes this reasoning precise.

Proof by Geometric Bounds

Write \(r_k=a_{k+1}/a_k\). First suppose \(L>0\). Choose any \(\varepsilon\) with \(0<\varepsilon<L\). By convergence of \(r_k\) to \(L\), there is an index \(N\) such that, for every \(k\geq N\),

L-\varepsilon<r_k<L+\varepsilon.

For \(n>N\), multiplying the ratios from \(k=N\) through \(k=n-1\) gives the exact identity

a_n=a_N\prod_{k=N}^{n-1}r_k.

There are \(n-N\) factors in this product. Since all quantities are positive, we can multiply the bounds without changing their direction, then take \(n\)th roots:

a_N(L-\varepsilon)^{n-N} < a_n < a_N(L+\varepsilon)^{n-N}, $$ $$ a_N^{1/n}(L-\varepsilon)^{(n-N)/n} < a_n^{1/n} < a_N^{1/n}(L+\varepsilon)^{(n-N)/n}.

The fixed positive factor \(a_N\) satisfies \(a_N^{1/n}\to 1\). Also, \((n-N)/n\to 1\). Therefore the left and right bounds tend to \(L-\varepsilon\) and \(L+\varepsilon\), respectively. In particular, the lower and upper limits of \(a_n^{1/n}\) lie between these two values. Because this holds for every \(\varepsilon\) with \(0<\varepsilon<L\), letting \(\varepsilon\) decrease to zero proves that \(a_n^{1/n}\to L\).

Now suppose \(L=0\). Given any \(\varepsilon>0\), convergence of \(r_k\) gives an \(N\) such that \(0<r_k<\varepsilon\) for every \(k\geq N\). The same product identity yields

0<a_n^{1/n}<a_N^{1/n}\varepsilon^{(n-N)/n} \qquad(n>N).

As \(n\to\infty\), the upper bound tends to \(\varepsilon\). Thus the nonnegative sequence \(a_n^{1/n}\) has upper limit at most \(\varepsilon\). This is true for every \(\varepsilon>0\), so \(a_n^{1/n}\to 0=L\). The two cases prove the theorem.

What the Theorem Says About the Tests

For a series with positive terms \(\sum a_n\), the ratio test uses the limit of \(a_{n+1}/a_n\), when it exists. The root test uses the limit of \(a_n^{1/n}\), when it exists. In either test, a limit below 1 implies convergence, and a limit above 1 implies divergence. A limit equal to 1 makes that test inconclusive.

The theorem shows that if the ratio limit exists and is finite, then the root limit exists and has exactly the same value. Consequently, whenever the ratio test reaches a decisive conclusion, the root test would reach the same conclusion if applied. In this sense, a ratio limit supplies enough information to determine the root limit as well.

“Stronger” needs careful interpretation here. The ratio-limit hypothesis is strong: it requires the successive ratios to settle to a finite value. Under that hypothesis, the root-limit conclusion follows automatically. But the theorem does not say that the ratio test settles more series than the root test. A root limit can exist even when successive ratios do not converge. Nor does the theorem make the tests decisive at the shared boundary \(L=1\).

Test-selection takeaway: When the ratio limit exists and is finite, it determines the root limit and both tests give the same classification. The ratio calculation may be more convenient, but a limit of 1 still requires a different test or argument.

Worked Examples

Worked Example: A Polynomial Factor Times Geometric Decay

Consider \(a_n=n^2/4^n\), with \(n\geq 1\). Each term is positive. Compute its successive ratio:

\frac{a_{n+1}}{a_n} = \frac{(n+1)^2/4^{n+1}}{n^2/4^n} = \frac{(n+1)^2}{n^2}\cdot\frac{4^n}{4^{n+1}} = \left(1+\frac1n\right)^2\cdot\frac14.

Since \((1+1/n)^2\to 1\), this ratio tends to \(1/4\). The theorem therefore gives

\lim_{n\to\infty}a_n^{1/n}=\frac14.

Indeed, directly taking the root gives \(a_n^{1/n}=n^{2/n}/4\), and \(n^{2/n}\to1\). Thus the root limit is \(1/4\), agreeing with the ratio limit. Since \(1/4<1\), both tests show that \(\sum_{n=1}^{\infty} n^2/4^n\) converges. Checking the ratio calculation in its original form confirms that no factor has been omitted: the powers contribute \(4^n/4^{n+1}=1/4\), and the polynomial terms contribute \((n+1)^2/n^2\).

Worked Example: Ratios That Approach Zero

Let \(a_n=1/n!\), for \(n\geq1\). Using \((n+1)!=(n+1)n!\), compute

\frac{a_{n+1}}{a_n} = \frac{1/(n+1)!}{1/n!} = \frac{n!}{(n+1)!} = \frac{1}{n+1}.

As \(n\to\infty\), this ratio tends to \(0\). The zero-limit case of the theorem gives

\lim_{n\to\infty}\left(\frac1{n!}\right)^{1/n}=0.

The ratio test has limit \(0<1\), so \(\sum_{n=1}^{\infty}1/n!\) converges. The root test reaches the same conclusion because its limit is also \(0<1\). As a check on the index, the ratio uses \(a_{n+1}=1/(n+1)!\); simplifying against \(a_n=1/n!\) leaves \(1/(n+1)\), not \(1/n\).

Worked Example: A Boundary Limit That Does Not Decide Convergence

Take \(a_n=1/n\). Its successive ratio is

\frac{a_{n+1}}{a_n} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} = \frac{1}{1+1/n} \longrightarrow 1.

Its \(n\)th root is

a_n^{1/n} = \left(\frac1n\right)^{1/n} = n^{-1/n} \longrightarrow 1.

Both tests therefore give the boundary value 1 and are inconclusive for \(\sum 1/n\). This is not a failure of the theorem: it correctly predicts that the two limits agree. It is a limitation of what those tests conclude at that value. The calculation is consistent at each step: \(n/(n+1)\to1\), and taking the \(n\)th root of \(1/n\) also gives a limit of 1.

A Pitfall: Treating the Boundary as Convergence

A common error is to interpret a ratio limit of 1 as evidence that the series converges. The ratio test does not say that. It says the test is inconclusive at 1. The same warning applies to the root test.

For comparison, take \(b_n=1/n^2\). Its ratio and root limits are

\frac{b_{n+1}}{b_n} = \frac{n^2}{(n+1)^2} = \left(\frac{n}{n+1}\right)^2 \longrightarrow 1, \qquad b_n^{1/n}=n^{-2/n}\longrightarrow 1.

The series \(\sum 1/n^2\) converges, whereas \(\sum 1/n\) diverges. Both have ratio limit 1 and root limit 1. The incorrect calculation would be: “The ratio limit is 1, so the series converges.” The correction is: “The ratio test is inconclusive; use another result or test.” A limit of 1 alone cannot distinguish these two outcomes.

How to Use the Relationship

When deciding whether to calculate ratios or roots, first ask what the term’s structure makes easiest to simplify. Ratios often cancel factorials and neighboring powers directly. Roots may be more transparent when a term is already expressed as a quantity raised to the index. If a finite ratio limit is available, the theorem guarantees that the corresponding root limit matches it.

Check positivity.
For this theorem, the terms must be positive so the ratios and real \(n\)th roots are defined as used.

Calculate the ratio limit.
Confirm that the index shift and cancellation are correct, then determine whether the limit is finite.

Transfer the limit to the roots.
If the ratio limit is finite, the theorem says the \(n\)th-root limit is the same.

Interpret the value carefully.
A limit below 1 gives convergence, a limit above 1 gives divergence, and a limit equal to 1 leaves these tests inconclusive.

For a series whose terms are not positive, ratio and root tests are commonly applied to absolute values. The same sequence theorem applies to \(|a_n|\) when the terms are nonzero from some index onward and the ratios \(|a_{n+1}|/|a_n|\) have a finite limit. The conclusions then concern absolute convergence or failure of the terms to approach zero, as appropriate.

Check Your Understanding

Use the theorem and the test conclusions carefully; distinguish a decisive limit from an inconclusive one.

If \(a_n>0\) and \(a_{n+1}/a_n\to 2/3\), what is the limit of \(a_n^{1/n}\), and what do the two tests conclude for \(\sum a_n\)?
Why does the proof compare \(a_n\) with products of \(L-\varepsilon\) and \(L+\varepsilon\)?
What extra step is needed in the proof when the ratio limit is \(L=0\)?
If the ratio limit is 1, what does the theorem imply about the root limit, and what does it not imply about the series?
Can the root limit exist when the successive-term ratios do not converge? Explain why the theorem does not rule this out.