Why Endpoints Need Their Own Tests
A power series has a radius of convergence, but that radius does not by itself determine what happens at the boundary. Inside the radius, the series converges absolutely; outside it, the series diverges. At a boundary point, however, either result is possible. The endpoint values must be substituted into the series and tested separately.
This separation matters because the two endpoints can produce different numerical series. A sign change at one endpoint may create an alternating series, while the other endpoint may produce a positive harmonic series. The same radius is consistent with both outcomes.
The radius property tells us that the series converges absolutely when \(|x-a|<R\) and diverges when \(|x-a|>R\). At \(|x-a|=R\), that property does not decide convergence. This tutorial focuses on those finite endpoints; finding the radius itself is a separate step.
What Substitution Gives at the Boundary
At the right endpoint \(x=a+R\), the quantity \(x-a\) equals \(R\), so substitution gives \(\sum_{n=0}^{\infty}c_nR^n\). At the left endpoint \(x=a-R\), \(x-a=-R\), so the resulting series is \(\sum_{n=0}^{\infty}c_n(-R)^n\). These may have different signs and different convergence behavior.
This substitution is the key justification for treating each endpoint as an ordinary series problem: once \(x\) is fixed at an endpoint, every term is a number, and the question is whether the resulting numerical series converges. The radius gives the boundary locations, not the endpoint answers.
Choosing a Test at an Endpoint
After substitution, identify the type of numerical series before choosing a test. The \(n\)th-term test says that if the terms of a series do not tend to zero, then the series diverges. A zero limit is necessary for convergence but is not enough to prove it.
Other common patterns include these:
- A geometric series \(\sum ar^n\) with a nonzero first term converges if \(|r|<1\) and diverges if \(|r|\ge 1\).
- A p-series \(\sum 1/n^p\) converges if \(p>1\) and diverges if \(p\le 1\).
- An alternating series \(\sum(-1)^n b_n\), with \(b_n\ge 0\), converges if \(b_n\) is eventually nonincreasing and tends to zero.
- For comparison, if \(0\le u_n\le v_n\) and \(\sum v_n\) converges, then \(\sum u_n\) converges. If \(0\le v_n\le u_n\) and \(\sum v_n\) diverges, then \(\sum u_n\) diverges.
When a series converges, also consider absolute convergence: test whether the series of absolute values converges. Convergence without absolute convergence is called conditional convergence. In endpoint work, this distinction is especially useful when substitution produces alternating signs.
Worked Example: One Endpoint Alternates and the Other Does Not
Worked Example: One Endpoint Alternates and the Other Does Not
Find the radius of convergence of \(\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}\), then test both endpoints.
For \(x\ne 2\), let \(u_n=\frac{(x-2)^n}{n}\). The ratio of successive absolute values is
As \(n\) tends to infinity, this ratio tends to \(|x-2|\). The ratio test gives absolute convergence when \(|x-2|<1\) and divergence when \(|x-2|>1\), so the radius is \(R=1\), centered at \(a=2\). The endpoints are \(1\) and \(3\).
At \(x=1\), \(x-2=-1\), and the series becomes
The values \(1/n\) decrease to zero: \(1/(n+1)<1/n\) for every \(n\ge1\), and \(\lim_{n\to\infty}1/n=0\). The alternating-series test therefore proves convergence. But the absolute-value series is \(\sum_{n=1}^{\infty}1/n\), the divergent p-series with \(p=1\). Thus this endpoint converges conditionally.
At \(x=3\), \(x-2=1\), and substitution gives
This is again the p-series with \(p=1\), so it diverges. The two endpoints have different outcomes even though each is exactly one unit from the center.
Check: Substituting \(x=1\) gives \((1-2)^n=(-1)^n\); substituting \(x=3\) gives \((3-2)^n=1\). These confirm the signs used in the endpoint series.
Worked Example: Absolute Convergence at Both Endpoints
Worked Example: Absolute Convergence at Both Endpoints
Test the endpoints of \(\sum_{n=1}^{\infty}\frac{(x-1)^n}{n(n+1)}\).
For \(x\ne1\), the ratio of successive absolute term values is
Its limit is \(|x-1|\), so the radius is \(R=1\). The endpoints are \(x=0\) and \(x=2\).
At \(x=0\), the series becomes \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n(n+1)}\). At \(x=2\), it becomes \(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\). For both, the absolute values give the same series. To test it, use the identity
The \(N\)th partial sum is therefore
As \(N\) tends to infinity, this partial sum tends to \(1\). Thus the absolute-value series converges, and both endpoints converge absolutely. The signed series at \(x=0\) must then converge as well, because absolute convergence implies convergence.
Check: At the left endpoint, \((0-1)^n=(-1)^n\); at the right endpoint, \((2-1)^n=1\). The absolute values at both endpoints are \(1/[n(n+1)]\), as used above.
Worked Example: The Terms Fail to Approach Zero
Worked Example: The Terms Fail to Approach Zero
Test the endpoints of the geometric power series \(\sum_{n=1}^{\infty}\left(\frac{x+1}{2}\right)^n\).
The ratio of consecutive terms in absolute value is \(\left|\frac{x+1}{2}\right|\), so the radius is \(R=2\) and the center is \(a=-1\). The endpoints are \(-3\) and \(1\).
At \(x=1\), \((x+1)/2=1\), so substitution gives \(\sum_{n=1}^{\infty}1\). Its terms equal \(1\), not zero, so the series diverges by the \(n\)th-term test.
At \(x=-3\), \((x+1)/2=-1\), so substitution gives \(\sum_{n=1}^{\infty}(-1)^n\). Its terms alternate between \(-1\) and \(1\), and therefore do not tend to zero. This series also diverges by the \(n\)th-term test. Equivalently, each is a geometric series with a nonzero first term and ratio of absolute value \(1\).
Check: The first term of the original series is \((x+1)/2\). At the endpoints it is \(1\) or \(-1\), so it is nonzero; the divergence conclusion is consistent with the geometric-series test.
A Common Mistake: Treating the Boundary as if It Were the Interior
A tempting but incorrect conclusion is that because both endpoints are the same distance from the center, they must have the same convergence behavior. The first worked example disproves this: at \(x=1\) it converges conditionally, while at \(x=3\) it diverges. The distance from the center identifies the boundary; it does not identify the behavior there.
Another careless shortcut is to say, “A geometric series with \(|r|\ge1\) diverges,” without checking its first term. For example, \(\sum_{n=0}^{\infty}0\cdot 2^n\) is the zero series and converges, despite the ratio having absolute value \(2\). The geometric divergence test applies when the first term is nonzero. In the third worked example, that condition was checked at both endpoints.
Why This Method Matters
Endpoint testing is where several familiar convergence ideas meet in power-series problems. Substitution can produce a p-series, a geometric series, an alternating series, or a series suited to comparison. The center and radius organize where to look, but the behavior at a boundary is governed by the particular coefficients and signs that remain after substitution.
A reliable solution therefore has two distinct stages. First, use the radius to locate the boundary values. Second, evaluate the series at each boundary value and justify its convergence or divergence with an appropriate test. If it converges, check whether it converges absolutely or only conditionally. This keeps the reasoning precise and prevents a correct radius calculation from being mistaken for a complete endpoint analysis.
Check Your Understanding
For each question, explain which test applies and why its conditions are met.
- Why does a finite positive radius fail to determine convergence at either endpoint?
- What numerical series results at \(x=a-R\) for \(\sum c_n(x-a)^n\)? What results at \(x=a+R\)?
- At an endpoint, how can an alternating series converge while failing to converge absolutely?
- What condition on the terms gives an immediate divergence conclusion by the \(n\)th-term test?
- Why must the first term be nonzero when applying the geometric-series divergence rule for \(|r|\ge1\)?
