From a Radius to a Complete Interval
A power series does not usually converge for every real input. Its terms may become small enough to add to a finite sum when \(x\) is near the center, but fail to do so farther away. The ratio test provides a systematic way to find the boundary between these regions.
Finding that boundary is not the last step. The ratio test generally identifies a radius, but does not decide whether either boundary point belongs to the interval of convergence. Each endpoint must be substituted into the original series and tested separately.
Power Series and the Ratio-Test Setup
A power series centered at \(c\) has the form \(\sum_{n=0}^{\infty} a_n(x-c)^n\), where \(c\) is a fixed real number and the \(a_n\) are coefficients. For a fixed value of \(x\), this is an ordinary infinite series whose terms are \(a_n(x-c)^n\).
Suppose the coefficients are eventually nonzero and the limit \(L=\lim_{n\to\infty}|a_{n+1}/a_n|\) exists. At a fixed \(x\ne c\), the ratio of the absolute values of successive terms is
As \(n\) increases, this ratio approaches \(L|x-c|\). By the ratio test, the series converges absolutely if this limit is less than \(1\), and diverges if it is greater than \(1\). When the limit equals \(1\), the ratio test gives no conclusion.
The restriction \(x\ne c\) matters: the displayed ratio is not defined at the center. At \(x=c\), all terms with \(n\ge 1\) are zero, so the series reduces to the single constant term \(a_0\) and converges. This conclusion handles the center separately rather than substituting it into a ratio that would be undefined.
Why the Radius Rule Works
The ratio test result can be seen directly from a geometric comparison. Fix \(x\ne c\), and let \(q=L|x-c|\), the limit of the successive absolute-term ratios. If \(q<1\), choose a number \(r\) with \(q<r<1\). Since the ratios approach \(q\), there is an index \(N\) such that every ratio after that index is less than \(r\). Thus the absolute terms after the \(N\)th term are bounded by a constant times \(r^k\), and the geometric series with ratio \(r\) converges. The power series therefore converges absolutely.
If \(q>1\), choose \(r\) with \(1<r<q\). Eventually, each absolute term is more than \(r\) times the preceding one. The terms then cannot approach zero, which is necessary for a series to converge. The power series diverges. When \(q=1\), neither argument applies: the terms might or might not approach zero, and even if they do, the series might still diverge. That is why the ratio test is inconclusive at the two boundary points.
For \(0<L<\infty\), the inequality on the right describes the open interval \((c-R,c+R)\), where \(R=1/L\). It identifies the interior, not automatically the full interval of convergence.
Worked Example: One Endpoint Converges Conditionally
Worked Example: One Endpoint Converges Conditionally
Find the radius and interval of convergence of \(\sum_{n=1}^{\infty}\frac{(x-2)^n}{n3^n}\).
Step 1: Identify the center and coefficients. The series is centered at \(c=2\), with \(a_n=1/(n3^n)\). These coefficients are nonzero for every \(n\ge 1\).
Step 2: Calculate the coefficient ratio.
Thus \(L=1/3\) and \(R=1/L=3\). For \(x\ne2\), the ratio of successive absolute terms approaches \(|x-2|/3\). The series converges absolutely if \(|x-2|<3\), or \(-1<x<5\), and diverges if \(|x-2|>3\).
Step 3: Test the left endpoint, \(x=-1\). Substitution gives
The resulting alternating harmonic series converges by the alternating-series test, while its absolute-value series \(\sum_{n=1}^{\infty}1/n\) diverges. Therefore the original series converges conditionally at \(x=-1\).
Step 4: Test the right endpoint, \(x=5\). Substitution gives
The harmonic series diverges, so \(x=5\) is not included. The interval of convergence is \([-1,5)\). As a check, the left bracket includes the conditionally convergent endpoint, while the right parenthesis excludes the divergent endpoint.
Worked Example: Both Endpoints Converge Absolutely
Worked Example: Both Endpoints Converge Absolutely
Find the radius and interval of convergence of \(\sum_{n=1}^{\infty}\frac{(x+1)^n}{n^2}\).
Step 1: Find the center and coefficient ratio. The center is \(c=-1\), and \(a_n=1/n^2\). Then
Therefore \(L=1\) and \(R=1\). For \(x\ne-1\), the ratio of successive absolute terms approaches \(|x+1|\). The series converges absolutely for \(|x+1|<1\), which gives \(-2<x<0\), and diverges for \(|x+1|>1\).
Step 2: Test the left endpoint, \(x=-2\).
The absolute-value series is \(\sum_{n=1}^{\infty}1/n^2\), which converges by the \(p\)-series test with \(p=2>1\). Hence the original series converges absolutely at \(x=-2\).
Step 3: Test the right endpoint, \(x=0\).
This is also a convergent \(p\)-series with \(p=2\), so \(x=0\) is included. The interval of convergence is \([-2,0]\). The endpoint checks, rather than the radius alone, justify both square brackets.
Worked Example: Neither Endpoint Converges
Worked Example: Neither Endpoint Converges
Find the radius and interval of convergence of \(\sum_{n=1}^{\infty}\frac{n(x-1)^n}{2^n}\).
Step 1: Find the center and coefficient ratio. The center is \(c=1\), and \(a_n=n/2^n\). Then
Thus \(L=1/2\) and \(R=2\). For \(x\ne1\), the successive absolute-term ratio approaches \(|x-1|/2\), so the series converges absolutely for \(|x-1|<2\), or \(-1<x<3\), and diverges for \(|x-1|>2\).
Step 2: Test the left endpoint, \(x=-1\).
The terms have absolute value \(n\), so they do not approach zero. The series diverges by the nth-term test.
Step 3: Test the right endpoint, \(x=3\).
These terms also do not approach zero, so the series diverges. Neither endpoint is included, and the interval of convergence is \((-1,3)\). This matches the endpoint tests: both boundaries fail the necessary condition that the terms approach zero.
A Reliable Endpoint Workflow
For an AP free-response question, keep the radius calculation and the endpoint work visibly separate. A clear solution makes it possible to see which conclusions come from the ratio test and which come from testing the boundary series.
Write the series in the form \(\sum a_n(x-c)^n\), and note its center \(c\).
For \(x\ne c\), find the limit of the successive absolute-term ratio and solve the resulting inequality.
Use \(R=1/L\) when \(0<L<\infty\). Do not yet decide whether the boundary values are included.
Choose an appropriate test for each resulting series; the two endpoints can behave differently.
Use a bracket for each convergent endpoint and a parenthesis for each divergent endpoint.
Common Mistake: Treating the Radius as a Closed Interval
A frequent error is to calculate \(R\) correctly and then write \([c-R,c+R]\) automatically. The radius only tells us where \(|x-c|\) is less than, equal to, or greater than \(R\). It does not imply convergence at equality.
For example, in the first worked example the radius is \(3\) and the center is \(2\). Writing \([-1,5]\) without checking the endpoints incorrectly includes \(x=5\). At that value, the series is \(\sum_{n=1}^{\infty}1/n\), which diverges. The corrected interval is \([-1,5)\): the left endpoint converges conditionally, and the right endpoint diverges.
Another error is to apply the ratio formula at \(x=c\). The formula for successive terms divides by a power of \(|x-c|\), so it requires \(x\ne c\). At the center, handle the series directly: all positive-power terms vanish, leaving only \(a_0\).
Check Your Understanding
For each question, show the endpoint substitutions when the series has a boundary to test.
- For \(\sum_{n=1}^{\infty}(x-4)^n/(n5^n)\), find the radius and test both endpoints.
- For \(\sum_{n=1}^{\infty}(x+2)^n/n^3\), determine the interval of convergence and classify each endpoint as absolutely convergent, conditionally convergent, or divergent.
- Why must the ratio calculation be handled separately at the center of a power series?
- What does a ratio-test limit of \(1\) establish about convergence?
- A power series has center \(c=3\) and radius \(R=2\). What are the two endpoint values that must be tested, and what determines whether each appears in the interval?
