Ratio And Root Tests at Radius Boundaries

Why These Tests Often Stop at the Boundary

The ratio and root tests are powerful tools for determining where a power series converges absolutely. At a boundary point, however, each test often returns the same value: \(1\). That value does not mean the series converges, and it does not mean the series diverges. It means that these tests do not decide the question.

The useful next step is to examine the series produced by substituting the endpoint and apply a test suited to that series. Depending on its terms, an alternating-series test, a p-series result, or the term test may settle the issue.

Ratio and Root Tests at an Endpoint

Write a power series centered at \(c\) as \(\sum_{n=0}^{\infty} a_n(x-c)^n\), with finite positive radius of convergence \(R\). Its boundary points are \(x=c-R\) and \(x=c+R\). At either point, the substituted terms are \(u_n=a_n(x-c)^n\). The ratio test and root test are applied to these terms, not to the original expression while leaving \(x\) unspecified.

Ratio test: If \(u_n\ne 0\) eventually and \(\lim_{n\to\infty}|u_{n+1}/u_n|=L\), the series \(\sum u_n\) converges absolutely when \(L<1\) and diverges when \(L>1\). When \(L=1\), the test gives no conclusion. If the limit does not exist, this version of the ratio test does not decide the series.

Root test: Let \(L=\limsup_{n\to\infty}\sqrt[n]{|u_n|}\). If \(L<1\), \(\sum u_n\) converges absolutely; if \(L>1\), it diverges. When \(L=1\), the test gives no conclusion. If the ordinary limit of the roots exists, it equals this limsup.

The reason the boundary value \(1\) occurs can be seen directly. At either endpoint, \(|x-c|=R\), so, whenever the relevant coefficients are nonzero,

\left|\frac{u_{n+1}}{u_n}\right| = R\left|\frac{a_{n+1}}{a_n}\right|, \qquad \sqrt[n]{|u_n|} = R\sqrt[n]{|a_n|}.

In particular, if the coefficient ratio tends to \(1/R\), the endpoint ratio limit is \(R(1/R)=1\). The radius formula also says that for a power series with finite positive radius \(R\), \(\limsup_{n\to\infty}\sqrt[n]{|a_n|}=1/R\). Thus at either endpoint the root-test limsup is \(R(1/R)=1\). These calculations explain why the tests commonly stall there; they do not determine whether the endpoint series itself converges.

Why a Limit of One Is Inconclusive

The ratio and root tests make decisions when the terms shrink at a sufficiently strong geometric rate, or when they fail to shrink fast enough. At the threshold value \(1\), they do not distinguish between series with different outcomes. For example, \(\sum 1/n\) diverges, whereas \(\sum 1/n^2\) converges, even though both have ratio and root test limits equal to \(1\).

The value \(1\) is therefore not a verdict. It is a signal to return to the endpoint series and use information about its particular terms. The following examples show the calculations and the appropriate next step.

Worked Example: The Harmonic and Alternating Harmonic Endpoints

Consider the power series

\sum_{n=1}^{\infty}\frac{x^n}{n}.

Its radius is \(1\), so test \(x=1\) and \(x=-1\) separately. At \(x=1\), the endpoint series is \(\sum_{n=1}^{\infty}1/n\). Its absolute terms are \(u_n=1/n\). The ratio is

\left|\frac{u_{n+1}}{u_n}\right| = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} \longrightarrow 1.

The root limit is also \(1\):

\sqrt[n]{|u_n|} = \sqrt[n]{\frac{1}{n}} = e^{-(\ln n)/n} \longrightarrow e^0=1.

Neither test decides convergence. This endpoint is the harmonic series, which diverges.

At \(x=-1\), substitution gives \(\sum_{n=1}^{\infty}(-1)^n/n\). The absolute values of its terms are still \(1/n\), so its ratio and root limits are the same \(1\) just calculated. The alternating-series test applies: \(1/n\) decreases because \(1/(n+1)<1/n\), and \(1/n\to0\). Therefore this endpoint series converges. Its absolute-value series is the divergent harmonic series, so the convergence is conditional.

The endpoint results differ even though the ratio and root limits are \(1\) at both points. The endpoint substitution, followed by a suitable test, is essential.

Worked Example: A Convergent p-Series at Both Endpoints

Consider

\sum_{n=1}^{\infty}\frac{x^n}{n^2}.

Its radius is \(1\). At \(x=1\), the endpoint series is \(\sum 1/n^2\). With \(u_n=1/n^2\), calculate the ratio:

\left|\frac{u_{n+1}}{u_n}\right| = \frac{1/(n+1)^2}{1/n^2} = \frac{n^2}{(n+1)^2} \longrightarrow 1.

The root limit is

\sqrt[n]{|u_n|} = \sqrt[n]{\frac{1}{n^2}} = e^{-2(\ln n)/n} \longrightarrow 1.

These tests do not decide the endpoint. Recognize instead the p-series \(\sum 1/n^p\), which converges for \(p>1\). Here \(p=2\), so the endpoint at \(x=1\) converges.

At \(x=-1\), the endpoint series is \(\sum (-1)^n/n^2\). Its absolute-value series is exactly \(\sum 1/n^2\), already identified as convergent. Thus the series at \(x=-1\) converges absolutely. The checks agree with the endpoint substitutions: at \(x=1\) every term is positive, and at \(x=-1\) the alternating signs disappear on taking absolute values.

A General Reason the Boundary Root Limit Is One

For a power series with finite positive radius \(R\), the root formula for the radius is \(1/R=\limsup_{n\to\infty}\sqrt[n]{|a_n|}\). At an endpoint \(x=c\pm R\), the absolute value of the \(n\)-th term is \(|a_n|R^n\). Consequently,

\limsup_{n\to\infty}\sqrt[n]{|a_n(x-c)^n|} = \limsup_{n\to\infty}\left(\sqrt[n]{|a_n|}R\right) = \frac{1}{R}R = 1.

This is a root-test limsup statement; it does not require the individual roots to have a limit. It explains why the root test cannot decide convergence at either endpoint of such a power series. The ratio test commonly produces \(1\) there as well when the coefficient ratio has the appropriate limit, but a ratio limit need not exist for every power series. In either case, apply the test to the actual endpoint series and respect what its hypotheses allow.

Worked Example: Boundary Terms That Do Not Approach Zero

Consider

\sum_{n=1}^{\infty}\frac{n}{n+1}x^n.

The coefficient ratio tends to \(1\), and the coefficient \(n\)-th roots tend to \(1\), so the radius is \(1\). At either endpoint \(x=1\) or \(x=-1\), the absolute value of the \(n\)-th term is \(u_n=n/(n+1)\).

The ratio test calculation is

\left|\frac{u_{n+1}}{u_n}\right| = \frac{(n+1)/(n+2)}{n/(n+1)} = \frac{(n+1)^2}{n(n+2)} = 1+\frac{1}{n(n+2)} \longrightarrow 1.

For the root test, \(1/2\le n/(n+1)<1\) for \(n\ge1\). Taking \(n\)-th roots gives

(1/2)^{1/n} \le \sqrt[n]{\frac{n}{n+1}} < 1.

Both bounds tend to \(1\), so the root limit is \(1\). Again, neither test decides the series.

Now use the term test. At \(x=1\), the terms \(n/(n+1)\) tend to \(1\), not \(0\). At \(x=-1\), the terms are \((-1)^n n/(n+1)\), whose absolute values also tend to \(1\); hence these terms do not tend to zero either. Since a convergent series must have terms tending to zero, both endpoint series diverge.

This conclusion checks the role of the term test: failure of the terms to approach zero proves divergence, whereas terms approaching zero would not, by itself, prove convergence.

A Practical Endpoint Procedure

At a boundary, keep the calculations and the final decision in separate stages. First substitute the endpoint. Then evaluate the ratio or root limit for the resulting terms. If the value is \(1\), do not infer either convergence or divergence from that test. Inspect the endpoint series and choose another valid test.

Substitute the endpoint.
Write the resulting numerical series explicitly, including any signs introduced by the endpoint.

Evaluate the ratio or root expression.
Use the endpoint terms \(u_n\). State the limit, and check that the test’s required limit exists when using the ratio test.

Interpret the result correctly.
A limit below \(1\) gives absolute convergence; a limit above \(1\) gives divergence. A value of \(1\) gives no conclusion.

Choose a test for the endpoint series.
Check whether its terms fail to approach zero, form a recognizable p-series, or satisfy the alternating-series test, among other possibilities.

Record each endpoint verdict separately.
Do not transfer the conclusion at one endpoint to the other; their substituted series can have different signs and convergence behavior.

A Common Mistake: Treating One as a Verdict

A frequent error is to calculate a ratio limit of \(1\) and then declare the endpoint divergent. For the series \(\sum (-1)^n/n\), that calculation is

\left|\frac{u_{n+1}}{u_n}\right| = \frac{n}{n+1} \longrightarrow 1.

The incorrect conclusion “therefore the series diverges” does not follow: the ratio test is inconclusive at \(1\). The correction is to test the endpoint series itself. Its terms alternate, their magnitudes \(1/n\) decrease to zero, and therefore the alternating-series test proves convergence. The series is not absolutely convergent because its absolute-value series is harmonic.

The reverse mistake is also possible: assuming that a root limit of \(1\) proves convergence. The series \(\sum 1/n\) has root limit \(1\) and diverges. A boundary limit of \(1\) is not evidence for either outcome; it only identifies the point at which that particular test has stopped being decisive.

Why This Distinction Matters

The radius of convergence describes convergence inside the radius and divergence outside it, but endpoint behavior needs separate analysis. Ratio and root tests help identify the geometric threshold, yet precisely at that threshold their strongest conclusions no longer apply. Recognizing this limitation prevents an incorrect interval of convergence and points to the next useful question: what kind of numerical series did endpoint substitution produce?

At one endpoint, alternating signs may allow convergence even when the absolute-value series diverges. At another, the terms might fail to approach zero. In other cases, the endpoint series may be a familiar p-series. The right decision comes from matching a test to the series, not from forcing a verdict out of a limit equal to \(1\).

Check Your Understanding

For each question, distinguish the endpoint calculation from the test’s conclusion.

For \(\sum x^n/n\), what series results at \(x=1\) and \(x=-1\), and why do the ratio and root tests fail to distinguish their convergence?
What does a ratio-test limit of \(1\) establish about convergence? What does it leave undecided?
For \(\sum x^n/n^2\), which test settles the endpoint at \(x=1\), and why does it also settle absolute convergence at \(x=-1\)?
Why does the term test prove divergence for \(\sum (n/(n+1))x^n\) at either endpoint?
What is the difference between the root test’s limsup being \(1\) and concluding that a series converges?