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Ratio, Root, and Test Selection · Tutorial 682 of 998

Fixed Bases Raised to the Index

Learn to classify series of powers from the limit of their bases, and recognize when the root test cannot decide the boundary case.

Advanced 9 min read

What You'll Learn

  • Compute the root-test quantity for terms of the form \((c_n)^n\).
  • Classify the series when the limiting base has modulus less than or greater than one.
  • Explain why a limiting base of modulus one is inconclusive.
  • Compare convergent and divergent series whose bases both approach the boundary.
  • Avoid confusing the limit of the bases with the limit of the terms.

Why the Base Limit Matters

A term raised to its own index can be difficult to judge by inspection. The root test is especially useful for such terms because taking the \(n\)th root cancels the exponent. When the base approaches a fixed number, the resulting limit often gives an immediate classification of the series.

The decisive distinction is whether the limiting base has modulus less than one, greater than one, or exactly one. The first two cases give definite conclusions; the boundary case does not.

Key idea: If \(a_n=(c_n)^n\), then \(\sqrt[n]{|a_n|}=|c_n|\). Thus the root test for \(\sum a_n\) reduces to examining the limit of the absolute values of the bases.

The Root-Test Calculation

For a series \(\sum a_n\), the root test examines the limit \(L=\lim_{n\to\infty}\sqrt[n]{|a_n|}\), provided that this limit exists. If \(L<1\), the series converges absolutely. If \(L>1\), the series diverges. If \(L=1\), the test gives no conclusion.

Now suppose the terms have the form \(a_n=(c_n)^n\), where \(c_n\) is a real number for each positive integer \(n\), and suppose \(c_n\to c\). Since absolute values remove any sign issue, we obtain

$$ \sqrt[n]{|a_n|} = \sqrt[n]{|(c_n)^n|} = |c_n|. $$

The absolute-value function is continuous, so \(c_n\to c\) implies \(|c_n|\to |c|\). Therefore the root-test limit is \(|c|\).

Theorem: Let \(a_n=(c_n)^n\), where \(c_n\to c\). If \(|c|<1\), then \(\sum a_n\) converges absolutely. If \(|c|>1\), then \(\sum a_n\) diverges. If \(|c|=1\), the root test is inconclusive, and the series must be studied by another argument.

The same classification includes a genuinely fixed base as a special case: if \(c_n=c\) for every \(n\), then \(a_n=c^n\), and the root-test limit is simply \(|c|\). When the bases vary with \(n\) but converge to \(c\), their eventual behavior is controlled by that limiting base unless it lies on the boundary \(|c|=1\).

Why the Classification Works

The calculation above identifies the root-test limit exactly. Here is why the resulting thresholds have their stated meaning. In the case \(|c|<1\), the root-test limit is below one, which is precisely the root test's condition for absolute convergence. In the case \(|c|>1\), the terms themselves fail to approach zero.

To see the latter directly, choose a number \(q\) with \(1<q<|c|\). Because \(|c_n|\to|c|\), there is an index \(N\) such that \(|c_n|>q\) whenever \(n\ge N\). Consequently,

$$ |a_n|=|c_n|^n>q^n\ge 1 \qquad(n\ge N). $$

Thus \(a_n\) cannot tend to zero. Since convergence of a series requires its terms to tend to zero, the series diverges. At \(|c|=1\), neither conclusion follows: terms can approach zero quickly enough for convergence, or fail to approach zero at all. The examples below show both possibilities.

1
Write the term clearly.
Identify \(a_n=(c_n)^n\), keeping the entire base inside the parentheses.
2
Take the absolute \(n\)th root.
Use \(\sqrt[n]{|a_n|}=|c_n|\).
3
Find the limiting base modulus.
If \(c_n\to c\), then the root-test limit is \(|c|\).
4
Apply the correct case.
A limit below one gives absolute convergence; a limit above one gives divergence; a limit equal to one requires another test or argument.

Worked Examples

Worked Example: A Limiting Base Below One

Classify the series \(\sum_{n=1}^{\infty}\left(\frac{2n+3}{6n+1}\right)^n\).

Set \(c_n=\frac{2n+3}{6n+1}\). Divide numerator and denominator by \(n\) to find its limit:

$$ c_n = \frac{2+3/n}{6+1/n} \longrightarrow \frac{2}{6} = \frac13. $$

For \(a_n=(c_n)^n\), the root-test quantity is

$$ \sqrt[n]{|a_n|} = \left|\frac{2n+3}{6n+1}\right| \longrightarrow \frac13. $$

Since \(\frac13<1\), the root test proves that the series converges absolutely. Check the classification by substituting the computed root limit into the criterion: \(L=\frac13\), so \(L<1\), exactly the absolute-convergence case.

Worked Example: A Limiting Base Greater Than One

Classify the series \(\sum_{n=1}^{\infty}\left(\frac{3n+2}{n+1}\right)^n\).

Here \(c_n=\frac{3n+2}{n+1}\), so

$$ c_n = \frac{3+2/n}{1+1/n} \longrightarrow 3. $$

Taking the \(n\)th root of the absolute value of the term gives

$$ \sqrt[n]{|a_n|} = \left|\frac{3n+2}{n+1}\right| \longrightarrow 3. $$

The root-test limit is greater than one, so the series diverges. This can also be checked directly: for every \(n\ge1\),

$$ \frac{3n+2}{n+1}>2 \quad\Longleftrightarrow\quad 3n+2>2n+2 \quad\Longleftrightarrow\quad n>0. $$

Thus \(a_n>2^n\ge2\) for every \(n\ge1\), and in particular the terms do not tend to zero. This confirms the divergence conclusion.

The Boundary: Same Base Limit, Different Outcomes

If \(c_n\to1\), the root-test limit equals one. That tells us the root test is inconclusive, not that the series diverges. In particular, the limit of the bases is not the limit of the terms: raising a base close to one to the \(n\)th power can produce very different term sizes.

Worked Example: Boundary Base with a Divergent Series

For \(n\ge1\), let \(c_n=n^{-1/n}\), and classify \(\sum_{n=1}^{\infty}(c_n)^n\).

First verify the limit of the base. Using logarithms,

$$ \ln(c_n) = -\frac{\ln n}{n} \longrightarrow 0, \qquad c_n = \exp\left(-\frac{\ln n}{n}\right) \longrightarrow 1. $$

The root-test quantity is therefore

$$ \sqrt[n]{|(c_n)^n|} = c_n = n^{-1/n} \longrightarrow 1. $$

The root test alone gives no conclusion. However, the terms simplify exactly:

$$ (c_n)^n = \left(n^{-1/n}\right)^n = n^{-1}. $$

The resulting series is the harmonic series \(\sum_{n=1}^{\infty}1/n\), which diverges. Check the distinction: the base limit is one, but the terms are \(1/n\), not one. Their series diverges even though the individual terms tend to zero.

Worked Example: Boundary Base with a Convergent Series

For \(n\ge1\), let \(c_n=n^{-2/n}\), and classify \(\sum_{n=1}^{\infty}(c_n)^n\).

Again, logarithms show that the bases approach one:

$$ \ln(c_n) = -\frac{2\ln n}{n} \longrightarrow 0, \qquad c_n = \exp\left(-\frac{2\ln n}{n}\right) \longrightarrow 1. $$

Taking the \(n\)th root gives

$$ \sqrt[n]{|(c_n)^n|} = c_n \longrightarrow 1. $$

So the root test is once again inconclusive. But this time the terms are

$$ (c_n)^n = \left(n^{-2/n}\right)^n = n^{-2}. $$

The series \(\sum_{n=1}^{\infty}1/n^2\) converges, so the given series converges. The calculation checks the result directly: both boundary examples have root limit one, but their terms reduce to different familiar series, with different outcomes.

A Common Mistake

A frequent error is to treat a root-test limit of one as if it were a divergence result. For example, after finding \(c_n\to1\), one might incorrectly write

$$ \lim_{n\to\infty}\sqrt[n]{|(c_n)^n|}=1 \quad\Longrightarrow\quad \sum_{n=1}^{\infty}(c_n)^n \text{ diverges}. $$

That implication is invalid: the root test makes no classification when its limit equals one. The correction is to inspect the terms or use another appropriate test. The preceding examples provide a concrete check: the same root limit \(1\) occurs for both \(\sum 1/n\), which diverges, and \(\sum 1/n^2\), which converges.

Another useful safeguard is to keep the order of operations clear. The root test takes the \(n\)th root of the absolute value of the whole term. For \(a_n=(c_n)^n\), that operation returns \(|c_n|\); it does not say that \(a_n\) itself approaches the limit of \(c_n\).

How to Use the Result

When a series contains powers with the index as exponent, first check whether the term can be written as \((c_n)^n\). If it can, identify the limiting base and compare its modulus with one. This is often much more efficient than trying to study the original terms directly.

Limiting baseRoot-test limitConclusion
\(|c|<1\)\(|c|<1\)Absolute convergence
\(|c|>1\)\(|c|>1\)Divergence
\(|c|=1\)\(1\)Inconclusive; examine the terms further

The boundary deserves special care. A limiting base of modulus one is a warning that the root test has reached its limit, not a verdict about the series. Exact simplification, comparison, or another suitable test may still settle the question.

Check Your Understanding

For each question, state the root-test limit and explain what it does or does not establish.

  1. If \(a_n=\left(\frac{5n+1}{8n+4}\right)^n\), what is the root-test limit, and what conclusion follows?
  2. If \(a_n=\left(\frac{4n-1}{n+2}\right)^n\), why must the series diverge?
  3. For \(c_n=n^{-3/n}\), find \(\lim c_n\) and simplify \((c_n)^n\). Does the root test classify the series?
  4. Why does a root-test limit of one fail to determine whether a series converges?
  5. What is the difference between the limit of the bases \(c_n\) and the limit of the terms \((c_n)^n\)?

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