Why Boundary Substitution Deserves Care
A power series can converge throughout the interior of its interval of convergence and behave differently at its two boundary points. The substitution itself is short, but errors in that step can change an alternating series into a nonalternating one, use the wrong endpoint, or lead to an unsupported conclusion about convergence.
For a power series centered at \(c\), the boundary points are determined by its center and radius \(R\). If \(R\) is finite and positive, those points are \(c-R\) and \(c+R\). To determine behavior there, substitute each value into the original series and then test the resulting numerical series. The two substitutions must be checked separately.
What Substitution Produces
Consider a power series \(\sum_{n=0}^{\infty}a_n(x-c)^n\) with finite positive radius of convergence \(R\). At the right endpoint, \(x=c+R\), the displacement from the center is \(x-c=R\). At the left endpoint, \(x=c-R\), it is \(x-c=-R\). Thus the endpoint series are obtained by replacing \((x-c)^n\) with \(R^n\) and \((-R)^n\), respectively.
This follows directly from substitution in each partial sum. If \(S_N(x)=\sum_{n=0}^{N}a_n(x-c)^n\), then \(S_N(c+R)=\sum_{n=0}^{N}a_nR^n\) and \(S_N(c-R)=\sum_{n=0}^{N}a_n(-R)^n\). Convergence at an endpoint means that the corresponding sequence of partial sums has a finite limit. The sign factor \((-1)^n\) at the left endpoint is part of the resulting terms; it must not be discarded.
Once the endpoint series has been written correctly, use an appropriate test on that numerical series. A necessary first check is the term test: if \(\sum u_n\) converges, then \(u_n\to 0\). Consequently, if the terms of a boundary series fail to approach zero, that series diverges. Passing the term test does not establish convergence; it only means that this particular reason for divergence has not ruled it out.
Worked Example: Keep the Left-Endpoint Sign
Worked Example: Keep the Left-Endpoint Sign
Determine the behavior at both boundary points of \(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(x-2)^n}{n}\). Its radius is \(1\), so its boundary points are \(2-1=1\) and \(2+1=3\).
At the left endpoint \(x=1\), the displacement is \(x-2=1-2=-1\). The \(n\)th term becomes
The endpoint series is therefore \(-\sum_{n=1}^{\infty}\frac{1}{n}\), which diverges: its partial sums are the negatives of the harmonic partial sums and decrease without bound.
At the right endpoint \(x=3\), the displacement is \(x-2=1\). The endpoint series is
Its terms alternate in sign, their magnitudes \(1/n\) decrease, and \(1/n\to0\). The alternating-series test gives convergence. The series is not absolutely convergent, because its absolute-value series is \(\sum_{n=1}^{\infty}1/n\), the divergent harmonic series. Thus the left endpoint diverges while the right endpoint converges conditionally.
Check: The substitutions used \(1-2=-1\) and \(3-2=1\). At the left endpoint the two sign factors multiply to \(-1\) for every \(n\), confirming that the result is not an alternating harmonic series.
Do Not Replace the Center with Zero
A common source of error is to treat the endpoint itself as the power-series displacement. The power is \((x-c)^n\), not \(x^n\), unless the center really is \(c=0\). With a shifted center, calculate \(x-c\) at each endpoint before raising it to the \(n\)th power. Also keep the radius distinct from the endpoint: \(R\) is the distance from the center, while the endpoint is \(c+R\) or \(c-R\).
Worked Example: Substitute Relative to the Center
Consider \(\sum_{n=1}^{\infty}\frac{(x-3)^n}{n2^n}\). Its radius is \(2\), so its endpoints are \(3-2=1\) and \(3+2=5\).
At \(x=1\), \(x-3=-2\). Substitution gives
The resulting series alternates, and its term magnitudes \(1/n\) decrease to zero, so it converges by the alternating-series test. Its absolute-value series is harmonic, so this convergence is conditional.
At \(x=5\), \(x-3=2\), so the terms become
The endpoint series is harmonic and diverges. As a check, the endpoint displacements are \(-2\) and \(2\), each with magnitude equal to the radius. Using \(1^n\) or \(5^n\) in place of \((x-3)^n\) would not represent the given power series.
Use the Term Test Before Other Tests
When an endpoint substitution produces a series with complicated signs or coefficients, calculate its general term and ask whether that term tends to zero. This simple check can settle divergence immediately. In particular, a boundary series does not converge merely because the original power series converges for every point strictly inside the radius.
The term test is a necessary condition, not a sufficient one. For instance, the terms of the harmonic series tend to zero, but the series still diverges. If the terms do tend to zero, choose a further test suited to the actual boundary series; an alternating-series test, comparison, or another appropriate test may be needed.
Worked Example: Reject a Boundary Series by Its Terms
Consider \(\sum_{n=1}^{\infty}n(x-1)^n\). Its radius is \(1\), so the endpoints are \(0\) and \(2\). For \(|x-1|<1\), the ratio of successive term magnitudes is \(\frac{n+1}{n}|x-1|\), which tends to \(|x-1|<1\); hence the series converges there.
At the right endpoint \(x=2\), each term is
These terms do not tend to zero, so the endpoint series diverges by the term test.
At the left endpoint \(x=0\), each term is
The magnitude of this term is \(n\), which also does not tend to zero. This endpoint series diverges as well. Alternating signs do not rescue a series whose terms fail the necessary term test.
Check: The endpoint terms are \(n\) and \(n(-1)^n\); their magnitudes are \(n\) at both boundaries, so neither sequence approaches zero.
Convergence at One Boundary Says Nothing Decisive About the Other
The two endpoint series differ by the factor \((-1)^n\), but that change can substantially affect convergence. It can create alternation, remove alternation, or leave the behavior unchanged, depending on the coefficients already present. Therefore, a convergence verdict at one endpoint cannot simply be transferred to the other.
The distinction is visible even for a power series centered at zero. At \(x=1\), its terms may all have the same sign; at \(x=-1\), substitution introduces \((-1)^n\). The resulting alternating series may converge even when its counterpart diverges, or the reverse can happen if the coefficients already alternate.
Worked Example: Test the Two Boundaries Independently
Consider \(\sum_{n=1}^{\infty}\frac{x^n}{n}\), which has radius \(1\). At \(x=1\), substitution gives \(\sum_{n=1}^{\infty}\frac{1}{n}\). This is the harmonic series and diverges.
At \(x=-1\), substitution gives
The magnitudes \(1/n\) decrease to zero, so this alternating series converges by the alternating-series test. Its absolute-value series is the divergent harmonic series, so it converges conditionally.
Check: Direct substitution yields \(1/n\) at the right endpoint and \((-1)^n/n\) at the left. The different verdicts follow from testing those two series, not from assuming one endpoint determines the other.
A Reliable Boundary-Checking Routine
A short routine helps prevent the most common errors. Write down the center and radius, calculate each endpoint, and substitute each one into the original power expression. Simplify the \(n\)th term before choosing a convergence test. Then verify that the test's conditions hold.
For center \(c\) and radius \(R\), use \(c-R\) and \(c+R\).
Evaluate \(x-c\) at the endpoint; do not confuse \(x\) with \(x-c\).
Keep factors such as \((-1)^n\) and combine all sign factors explicitly.
First check whether the terms tend to zero; then use a suitable test if needed.
A result at one endpoint does not establish the result at the other.
Common Mistakes and Their Corrections
| Mistake | Why it fails | Correction |
|---|---|---|
| Using \(x^n\) when the series contains \((x-c)^n\) | The powers measure displacement from the center. | Substitute the endpoint into \(x-c\) first. |
| Dropping \((-1)^n\) at the left endpoint | The left displacement is \(-R\), whose \(n\)th power carries the sign. | Write \((-R)^n=(-1)^nR^n\). |
| Calling a series alternating without simplifying its full term | Coefficient signs may cancel or combine with the endpoint sign. | Multiply all sign factors before identifying the pattern. |
| Skipping the term test | A series with terms that do not approach zero cannot converge. | Calculate the general boundary term and its limit first. |
| Copying one endpoint's verdict to the other | The two boundary series can have different signs and convergence behavior. | Substitute and test each endpoint separately. |
Why the Details Matter
Endpoint conclusions depend on exact algebra, not on the appearance of the original power series. Losing a negative sign can reverse whether terms alternate. Using the wrong displacement can produce an entirely different series. Skipping the term test can obscure an immediate divergence, while relying on the opposite endpoint can turn a valid conclusion into an unsupported one.
A dependable analysis therefore separates two tasks: first obtain the correct numerical series by substitution; then establish its convergence or divergence with a test whose conditions are satisfied. This makes each boundary verdict independently checkable.
Check Your Understanding
For each question, show the substitution or reasoning that supports your answer.
- For a series centered at \(c\) with radius \(R\), what values of \(x-c\) occur at the left and right boundaries?
- At the left endpoint, why must the factor \((-1)^n\) be retained when simplifying \((x-c)^n\)?
- What does the term test let you conclude if the terms of a boundary series do not approach zero? What does it not let you conclude if they do approach zero?
- For \(\sum_{n=1}^{\infty}(-1)^{n-1}(x-2)^n/n\), what series results at \(x=1\), and why does it diverge?
- Why does convergence at one boundary point fail to determine convergence at the other?