What the Radius Tells Us—and What It Does Not
A power series has a radius of convergence that separates points where the series converges from points where it diverges. The radius gives a strong conclusion for every point strictly inside or strictly outside its boundary. At a boundary point, however, the standard root and ratio tests usually give the value \(1\), which is inconclusive.
This distinction matters when using a radius to describe convergence: the radius identifies the distance to the boundary, but it does not by itself decide whether the boundary points are included. The reason comes directly from applying the tests to the terms of the series.
The Root Test Applied to a Power Series
Consider a power series centered at \(c\), with coefficients \(a_n\):
For a fixed \(x\), its \(n\)th term is \(u_n=a_n(x-c)^n\). The root test examines the limit superior of the \(n\)th roots of the absolute values of these terms. The Cauchy–Hadamard formula defines the radius \(R\) using the coefficients:
This formula includes the conventions \(1/0=\infty\) and \(1/\infty=0\), allowing \(R\) to be infinite or zero. In what follows, assume \(0<R<\infty\). Since \(|x-c|\) does not depend on \(n\), taking \(n\)th roots of the series terms gives
The root test says a series of terms \(u_n\) converges absolutely if \(\limsup |u_n|^{1/n}<1\), and diverges if \(\limsup |u_n|^{1/n}>1\). When that value is \(1\), the test gives no conclusion. The theorem for power series follows by comparing the value \(|x-c|/R\) with \(1\).
Why the Inside and Outside Conclusions Follow
The root-test calculation captures the main idea, but the conclusions can also be seen through geometric bounds. This makes clear why strict inequality is needed and why equality changes the situation.
Let \(r=|x-c|<R\). Then \(r/R<1\), so choose a number \(q\) with \(r/R<q<1\).
Because \(\limsup |a_n|^{1/n}=1/R\) and \(1/R<q/r\) when \(r>0\), the definition of limit superior gives \(|a_n|^{1/n}<q/r\) for all sufficiently large \(n\). Thus \(|a_n(x-c)^n|<q^n\) for those \(n\).
The tail is bounded by a constant initial segment plus \(\sum q^n\), which converges because \(0<q<1\). Hence the power series converges absolutely. If \(r=0\), all terms with \(n\ge1\) are zero, so convergence is immediate.
If \(r>R\), then \(r/R>1\). The root test gives \(\limsup |a_n(x-c)^n|^{1/n}>1\), so the terms cannot tend to zero. Since convergence requires its terms to tend to zero, the series diverges.
At \(r=R\), the inside argument loses its geometric decay: there is no number \(q<1\) that can be chosen with \(r/R<q\). The outside argument also fails: the root-test value is not greater than \(1\), so it does not show that the terms fail to approach zero. Equality is exactly where these arguments stop deciding the question.
Worked Example: A Geometric Series Inside and Outside Its Radius
Worked Example: A Geometric Series Inside and Outside Its Radius
Find the radius of convergence of \(\sum_{n=0}^{\infty}2^n(x-1)^n\), then test \(x=1.25\) and \(x=2\) using the root test.
Here \(a_n=2^n\). The coefficient roots are \(|a_n|^{1/n}=2\) for \(n\ge1\), so
At \(x=1.25\), the distance from the center is \(|1.25-1|=0.25\). Thus \(|x-1|/R=0.25/0.5=0.5<1\), so the series converges absolutely. Indeed, its terms are \(2^n(0.25)^n=(0.5)^n\), a geometric series with sum \(1/(1-0.5)=2\). Checking the ratio of consecutive terms gives \(0.5\), as expected.
At \(x=2\), the distance is \(|2-1|=1\), so \(|x-1|/R=1/0.5=2>1\). The terms are \(2^n\), which do not tend to zero; the series diverges. The root-test value is \(2\), confirming divergence. These checks agree with the rule: strict interior points converge, and strict exterior points diverge.
What Happens at the Boundary
If \(|x-c|=R\), then the root-test calculation gives
This is not a convergence result or a divergence result. It means only that the root test does not decide the series at that point. Different coefficient sequences can produce different behavior at the boundary, even when they have the same radius.
For the ratio test, suppose the coefficient ratios \(|a_{n+1}/a_n|\) are defined for all sufficiently large \(n\), and have a finite limit \(L>0\). Then the ratio test applied at a fixed \(x\) examines
Under these assumptions, the radius is \(R=1/L\). At a boundary point, \(|x-c|=R=1/L\), so the ratio-test limit is \(L(1/L)=1\). The ratio test also gives no conclusion. If the coefficient ratios do not have a suitable limit, the ratio test may fail to decide the radius or may not apply in this form at all.
Worked Example: A Boundary Series That Converges
Worked Example: A Boundary Series That Converges
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n(n+1)}\). Find its radius and examine its value at the boundary point \(x=1\).
The coefficient is \(a_n=1/[n(n+1)]\). Its \(n\)th root tends to \(1\), because
Therefore \(1/R=1\), so \(R=1\). At \(x=1\), the terms are \(1/[n(n+1)]\). Use the identity \(1/[n(n+1)]=1/n-1/(n+1)\). The \(N\)th partial sum is
As \(N\to\infty\), this tends to \(1\), so the boundary series converges. The ratio of consecutive term magnitudes at \(x=1\) is
Thus the ratio test gives its inconclusive value even though the series converges. The root test does too, since the \(n\)th roots of its terms tend to \(1\).
Worked Example: A Boundary Series That Diverges
Worked Example: A Boundary Series That Diverges
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n}\). Find its radius and determine what happens at \(x=1\).
Here \(a_n=1/n\), and \(n^{1/n}\to1\), so \(|a_n|^{1/n}\to1\). Hence \(R=1\). At \(x=1\), the series becomes \(\sum_{n=1}^{\infty}1/n\). To verify divergence, group terms in blocks whose lengths double. In the block from \(2^k+1\) through \(2^{k+1}\), there are \(2^k\) terms, and each is at least \(1/2^{k+1}\). Therefore the block sum is at least
There are arbitrarily many such blocks, each contributing at least \(1/2\), so the partial sums are unbounded and the series diverges. Nevertheless, its root-test value at \(x=1\) is \(\lim (1/n)^{1/n}=1\). Its ratio-test limit is also
Both tests are inconclusive at this boundary point. Comparing this example with the preceding one shows that the same radius and the same inconclusive test values can occur with different boundary outcomes.
A Common Mistake
A frequent error is to treat the radius as if it included its boundary automatically. For example, the series \(\sum_{n=0}^{\infty}x^n\) has radius \(R=1\). It is tempting to claim that it converges whenever \(|x|\le1\), but at \(x=1\) it becomes
whose partial sums are \(1,2,3,\ldots\) and do not converge. The root-test value at \(x=1\) is \(1\), so the test gives no verdict—not permission to include the point. The correction is to use the radius only for the strict inequalities \(|x-c|<R\) and \(|x-c|>R\), then analyze boundary points separately.
Why Boundary Analysis Matters
The radius provides a reliable map of the interior and exterior of the convergence region. For a real power series with finite positive radius \(R\), every point strictly between \(c-R\) and \(c+R\) gives absolute convergence, and every real point beyond those limits gives divergence. The two boundary points are not classified by this information alone.
This is why a radius and a convergence interval are not interchangeable. The radius specifies a distance; deciding whether an endpoint belongs to the interval requires checking the series at that endpoint. The root and ratio tests can guide that later work, but when they return \(1\), another argument is needed.
Check Your Understanding
Answer each question by using the test conditions carefully.
- For a power series with \(R=3\), what does the root-test calculation give when \(|x-c|=2\), and what conclusion follows?
- What value does the root test give when \(|x-c|=R\)? Why does that value not decide convergence?
- If the coefficient-ratio limit is \(L=2\), what is the radius, and what ratio-test limit occurs at a boundary point?
- Why does a convergent power series at one boundary point not imply convergence at the other boundary point?
- State the error in claiming that every power series converges at all points satisfying \(|x-c|\le R\).