Why Polynomial Decay Defeats Both Tests

When Two Tests Reach the Same Boundary

The ratio test and the root test are often effective when a series contains factorials, exponentials, or terms raised to the index. Polynomial decay behaves differently. For a term such as \(1/n^p\), with \(p>0\), each test reaches the value one—the precise boundary at which neither test decides convergence.

That result is not a verdict about the series. Some series with these terms converge, and others diverge. The important skill is to calculate the limits correctly, interpret the boundary result accurately, and then use a suitable fact to decide the series when one is available.

Central takeaway: If the ratio-test limit or root-test limit equals one, that test is inconclusive. It does not prove convergence, and it does not prove divergence.

What the Tests Say at the Boundary

For a series \(\sum_{n=1}^{\infty} a_n\), the ratio test examines the limit of \(\lvert a_{n+1}/a_n\rvert\), when the terms needed for the quotient are nonzero. If this limit exists and is less than one, the series converges absolutely. If it exists and is greater than one, the series diverges. If it equals one, the test gives no conclusion.

The root test examines the limit of \(\sqrt[n]{\lvert a_n\rvert}\). A limit less than one implies absolute convergence; a limit greater than one implies divergence. A limit equal to one is again inconclusive. These outcomes concern what can be established from the test limit alone; other information about the series may still settle its behavior.

Definition: For a series with nonzero terms, its ratio-test limit is \(\lim_{n\to\infty}\lvert a_{n+1}/a_n\rvert\), if that limit exists. Its root-test limit is \(\lim_{n\to\infty}\sqrt[n]{\lvert a_n\rvert}\), if that limit exists. When either limit equals one, that test is inconclusive.

Deriving the Limits for Polynomial Decay

Consider \(a_n=n^{-p}=1/n^p\), where \(p>0\) is fixed. The ratio compares consecutive terms. The root takes the \(n\)th root of the term itself. Although both calculations involve \(p\), both limits turn out to be independent of its value.

\frac{a_{n+1}}{a_n} = \frac{(n+1)^{-p}}{n^{-p}} = \left(\frac{n}{n+1}\right)^p = \left(\frac{1}{1+1/n}\right)^p \longrightarrow 1^p=1.

For the root limit, rewrite the root using the exponential and logarithm:

\sqrt[n]{a_n} = (n^{-p})^{1/n} = n^{-p/n} = \exp\left(-p\frac{\ln n}{n}\right).

To see why the exponent tends to zero, for \(x\geq 1\) use the integral definition of the logarithm. Since \(1/t\leq 1/\sqrt{t}\) for \(t\geq 1\),

0\leq \ln x = \int_1^x \frac{1}{t}\,dt \leq \int_1^x \frac{1}{\sqrt{t}}\,dt = 2(\sqrt{x}-1) \leq 2\sqrt{x}.

Dividing by \(x>0\) gives \(0\leq \ln x/x\leq 2/\sqrt{x}\). The upper bound tends to zero, so \(\ln n/n\to 0\). Since \(p\) is fixed, \(-p\ln n/n\to 0\). Continuity of the exponential then gives the root limit.

\lim_{n\to\infty}\sqrt[n]{n^{-p}} = \exp(0) = 1.

Result: For every fixed \(p>0\), the terms \(a_n=n^{-p}\) give ratio-test limit one and root-test limit one. Neither test decides whether \(\sum_{n=1}^{\infty}1/n^p\) converges.

Worked Example: A Convergent Series with Both Limits One

Worked Example: The Series with Terms \(1/n^2\)

Consider \(\sum_{n=1}^{\infty}1/n^2\). Its terms are positive, and \(a_n=1/n^2\), so \(a_{n+1}=1/(n+1)^2\). The ratio is

\frac{a_{n+1}}{a_n} = \frac{1/(n+1)^2}{1/n^2} = \frac{n^2}{(n+1)^2} = \left(\frac{n}{n+1}\right)^2 \longrightarrow 1.

The root limit is

\sqrt[n]{a_n} = \sqrt[n]{n^{-2}} = n^{-2/n} = \exp\left(-2\frac{\ln n}{n}\right) \longrightarrow \exp(0)=1.

Both tests are inconclusive. The p-series criterion states that \(\sum_{n=1}^{\infty}1/n^p\) converges exactly when \(p>1\). Here \(p=2>1\), so the series converges. This is a convergence conclusion from the p-series criterion, not from either test limit.

Check: the calculated ratio and root limits are each exactly at the tests’ inconclusive boundary, while the separate p-series criterion classifies the series as convergent. The two conclusions are consistent because the tests did not claim divergence.

Worked Example: A Divergent Series with Both Limits One

Worked Example: The Harmonic Series

Now consider \(\sum_{n=1}^{\infty}1/n\), with \(a_n=1/n\) and \(a_{n+1}=1/(n+1)\). The ratio is

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

The root limit is

\sqrt[n]{a_n} = \sqrt[n]{n^{-1}} = n^{-1/n} = \exp\left(-\frac{\ln n}{n}\right) \longrightarrow \exp(0)=1.

The ratio and root tests again give no conclusion. The p-series criterion gives the answer: here \(p=1\), and the p-series diverges when \(p\leq 1\). Thus the harmonic series diverges.

Check: its two test limits are the same as for the convergent series in the preceding example, but the p-series criterion classifies this one as divergent. This contrast shows directly why the boundary value one cannot, by itself, decide the outcome.

Worked Example: A Different Convergent Power

Worked Example: Terms \(1/n^{3/2}\)

Consider \(\sum_{n=1}^{\infty}1/n^{3/2}\). Here \(a_n=n^{-3/2}\), so

\frac{a_{n+1}}{a_n} = \frac{(n+1)^{-3/2}}{n^{-3/2}} = \left(\frac{n}{n+1}\right)^{3/2} \longrightarrow 1.

Taking the \(n\)th root gives

\sqrt[n]{a_n} = (n^{-3/2})^{1/n} = n^{-3/(2n)} = \exp\left(-\frac{3}{2}\frac{\ln n}{n}\right) \longrightarrow \exp(0)=1.

Both tests are inconclusive. Since \(p=3/2>1\), the p-series criterion says that this series converges. As a check, the ratio and root calculations both return one, and neither test has been used to assert convergence; that assertion comes from the stated criterion.

Why the Same Limits Can Hide Different Outcomes

The ratio measures the relative change from one term to the next. For \(1/n^p\), consecutive terms become close in relative size: their ratio approaches one. The root test also approaches one because taking an \(n\)th root weakens the effect of the fixed power \(p\). In the root calculation, this appears in the factor \(\ln n/n\), which tends to zero.

These limits describe aspects of how the terms change; they do not record enough information to decide the sum at this boundary. In particular, the value of \(p\) determines whether the p-series converges, even though that value disappears from both limiting answers. The p-series criterion is therefore useful here: it distinguishes cases the ratio and root tests leave unresolved.

Use the conclusion precisely: “The ratio test is inconclusive because its limit is one” is a complete and correct conclusion about that test. It is not the same statement as “the series diverges” or “the series converges.”

A Common Mistake: Calling One a Divergence Result

A frequent error is to treat a ratio limit of one as if it were greater than one. For example, a mistaken calculation might say:

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1 \quad\Longrightarrow\quad \text{the series diverges}.

That implication is not part of the ratio test. Divergence is established by the test when its limit is greater than one, not when it equals one. The correction is:

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1 \quad\Longrightarrow\quad \text{the ratio test gives no conclusion}.

There is a related root-calculation error. Since \(a_n=n^{-p}\), it is incorrect to write \(\sqrt[n]{a_n}=n^{-p}\). The exponent must also be divided by \(n\):

\sqrt[n]{n^{-p}}=(n^{-p})^{1/n}=n^{-p/n},

which tends to one, not to \(n^{-p}\). Writing the exponent explicitly is a reliable way to avoid this mistake.

Choosing What to Do Next

When a ratio or root calculation gives one for terms with polynomial decay, stop before assigning a convergence verdict. Record the test as inconclusive, then look for a separate result that fits the series. For the exact p-series \(\sum 1/n^p\), the criterion is direct: it converges for \(p>1\) and diverges for \(0<p\leq 1\). The examples above illustrate both sides of that threshold.

The broader lesson is about interpreting a test, not forcing it to answer a question outside its conditions. A test that returns a boundary value has still provided accurate information; it has simply not distinguished between possible outcomes. Recognizing that limitation is part of applying convergence tests correctly.

Check Your Understanding

For each question, distinguish carefully between what the ratio or root test establishes and what requires another result.

For \(a_n=n^{-4}\), calculate the ratio-test limit and the root-test limit.
Why does a root-test limit of one fail to establish either convergence or divergence?
For the series \(\sum_{n=1}^{\infty}1/n^{2/3}\), what do the ratio and root tests conclude? What does the p-series criterion conclude?
Identify the error in the statement “the ratio limit is one, so the series diverges.”
Rewrite \(\sqrt[n]{n^{-p}}\) correctly and state its limit for fixed \(p>0\).

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