How Coefficient Size Predicts Endpoint Behavior
A power series can converge throughout the interior of its interval of convergence while behaving differently at its two endpoints. At a boundary point, the powers of the centered variable have fixed magnitude, so the remaining coefficient size and sign pattern become decisive.
Suppose a real power series is centered at \(c\) and has a finite positive radius of convergence \(R\). At the right endpoint \(x=c+R\), its terms are \(a_nR^n\). At the left endpoint \(x=c-R\), they are \((-1)^n a_nR^n\). Thus, even if the coefficient magnitudes are the same at both ends, substitution can introduce alternating signs at the left endpoint.
To focus on the endpoint terms, define \(u_n=a_nR^n\). We will study the case in which these scaled coefficients have a power-law asymptotic form. If \(R=1\), then \(u_n=a_n\), so the condition is directly about the original coefficients.
We take \(C\) and \(u_n\) to be real. A finite number of initial terms never changes whether a series converges, so the endpoint analysis can focus on the eventual behavior of its terms.
The Basic Endpoint Verdicts
The comparison is with the \(p\)-series \(\sum 1/n^p\): it converges for \(p>1\) and diverges for \(p\le 1\). At the right endpoint, the terms are \(u_n\), which eventually have a fixed sign. At the left endpoint, the terms are \((-1)^nu_n\), which eventually alternate in sign.
| Exponent | Right endpoint: \(\sum u_n\) | Left endpoint: \(\sum (-1)^n u_n\) |
|---|---|---|
| \(p>1\) | Absolutely convergent | Absolutely convergent |
| \(0<p\le 1\) | Divergent | Not decided by the asymptotic estimate alone; converges if the alternating-series conditions hold, and then is conditional |
| \(p\le 0\) | Diverges by the term test | Diverges by the term test |
Here is why the first and third rows follow directly. Since \(|u_n|\sim |C|/n^p\), for \(p>1\) limit comparison with the convergent \(p\)-series gives \(\sum |u_n|<\infty\). The left endpoint has the same term magnitudes, so it too converges absolutely. If \(p\le0\), then \(|u_n|\sim |C|n^{-p}\) does not tend to zero. The terms at neither endpoint tend to zero, so the necessary term test for convergence fails.
For \(0<p\le1\), the right endpoint diverges: its terms eventually have one sign, and their magnitudes are comparable to \(1/n^p\), whose series diverges. At the left endpoint, however, cancellation may matter. If the magnitudes eventually decrease to zero, the alternating-series test gives convergence. Because \(\sum |u_n|\) still diverges for \(0<p\le1\), that convergence is conditional, not absolute.
Why Substitution Creates the Sign Pattern
For a series centered at \(c\), the term with index \(n\) is \(a_n(x-c)^n\). Substituting the two boundary values gives
The factor \((-1)^n\) comes from the substitution, not from an assumption that the original coefficients alternate. Since \(u_n\) eventually has the fixed sign of \(C\), the left-endpoint terms alternate after at most a finite number of terms.
The alternating-series test applies when the terms can be written, apart from an overall sign and finitely many initial terms, as alternating positive magnitudes \(v_n\) with \(v_n\) nonincreasing and \(v_n\to0\). Its conclusion is convergence. The test does not require the magnitudes to be exactly \(1/n^p\), but an asymptotic equivalence to \(1/n^p\) does not establish that they are nonincreasing.
A Short Justification of the Alternating-Series Test
For completeness, suppose \(v_n\ge0\), \(v_n\) is nonincreasing, and \(v_n\to0\). Consider the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}v_n\). Its even partial sums increase because each added pair contributes \(v_{2k-1}-v_{2k}\ge0\). They are bounded above by the first partial sum \(v_1\), since the remaining terms can be grouped into nonpositive pairs after that partial sum. Thus the even partial sums converge.
The odd partial sums decrease, are bounded below, and also converge. The difference between the \(N\)th odd and preceding even partial sums is \(v_{2N+1}\), which tends to zero. Therefore the two sequences of partial sums have the same limit, and the full sequence of partial sums converges. This argument explains why both the alternating sign pattern and the decrease-to-zero condition matter.
Worked Endpoint Examples
Worked Example: A Power Larger Than One
Consider the power series \(\sum_{n=1}^{\infty} x^n/(2^n n^{3/2})\). Its radius of convergence is \(2\), and at either endpoint the scaled coefficient magnitude is \(1/n^{3/2}\).
At \(x=2\), each term is
The series \(\sum 1/n^{3/2}\) converges because \(3/2>1\). Thus the right endpoint converges.
At \(x=-2\), each term is
The absolute values again sum to \(\sum 1/n^{3/2}\), so the left endpoint also converges absolutely. As a check, the absolute term series is identical at both endpoints and is a convergent \(p\)-series with \(p=3/2\).
Worked Example: Harmonic Size at the Two Endpoints
Consider \(\sum_{n=1}^{\infty}x^n/n\), which has radius of convergence \(1\). Here the endpoint-scaled coefficients are \(u_n=1/n\), so \(C=1\) and \(p=1\).
At \(x=1\), the series becomes
This is the divergent harmonic series. At \(x=-1\), substitution gives
The magnitudes \(1/n\) decrease to zero, so the alternating-series test proves convergence. The absolute-value series is the divergent harmonic series, so the left endpoint is conditionally convergent. These two verdicts show how substitution changes endpoint behavior even though the term magnitudes match.
Why Asymptotic Size Alone Can Be Insufficient
A common error is to reason that \(u_n\sim C/n^p\) with \(p>0\) automatically makes the left-endpoint series converge by alternation. The alternating-series test needs more than terms tending to zero: it also needs the magnitudes to decrease (or some other sufficient argument for convergence). The following example shows that the missing condition cannot simply be inferred from asymptotic equivalence.
Worked Example: An Alternating Endpoint That Still Diverges
Define positive coefficients by
and, for integers \(k\ge2\), define
These definitions cover every positive integer index: the separately specified terms are \(b_1,b_2\), and the formulas for \(k\ge2\) define \(b_3,b_4,b_5,b_6,\ldots\). For \(k\ge2\), \(\log(k+1)>1\), so all displayed coefficients are positive.
For odd indices, the ratio of \(b_{2k-1}\) to \(1/(2k-1)\) is \(1-1/\log(k+1)\), which tends to \(1\). For even indices, the ratio of \(b_{2k}\) to \(1/(2k)\) is \(1+1/\log(k+1)\), which also tends to \(1\). Therefore \(b_n\sim1/n\).
The right-endpoint series \(\sum b_n\) diverges by limit comparison with the harmonic series. Now group consecutive terms in the left-endpoint series \(\sum(-1)^n b_n\). Each pair is
Writing \(L=\log(k+1)\), calculate the pair exactly:
The parenthesized sum is greater than \(1/k\), and \(1/[2k(2k-1)]<1/(2k^2)\). For all sufficiently large \(k\), \(\log(k+1)\le k\), so
The series \(\sum 1/[k\log(k+1)]\) diverges; for example, its divergence follows from the integral comparison with \(\int dx/(x\log x)=\log(\log x)\). Hence the sums of these consecutive pairs grow without bound. The left-endpoint series diverges as well. We have \(b_n\sim1/n\), yet alternation alone does not yield convergence. The explicitly specified value \(b_2=1\) ensures the sequence is defined at every index; changing finitely many initial terms would not affect either divergence conclusion.
Practical Endpoint Analysis
When an asymptotic coefficient estimate is available, separate the tasks. First find the actual terms at each endpoint, including any powers of the radius and the factor \((-1)^n\). Then use the exponent to decide absolute convergence or failure of the term test. Only after that should you use an alternating-series argument, and only if its conditions are verified.
For radius \(R\), use \(u_n=a_nR^n\), so the endpoint terms are \(u_n\) and \((-1)^nu_n\).
If \(|u_n|\sim |C|/n^p\), compare with the \(p\)-series: absolute convergence holds for \(p>1\), while terms fail to tend to zero for \(p\le0\).
For \(0<p\le1\), the right endpoint diverges. At the left endpoint, verify a valid alternating-series condition rather than inferring convergence from asymptotic size alone.
The main lesson is that endpoint verdicts depend on both magnitude and sign structure. Asymptotic coefficients usually settle absolute convergence and the nonalternating boundary. When substitution creates alternating signs, the additional behavior of the magnitudes can decide whether cancellation is sufficient.
Check Your Understanding
Use endpoint-scaled coefficients in your reasoning, and state any extra condition needed for an alternating-series conclusion.
- If \(u_n\sim 4/n^{5/4}\), what can be concluded at both endpoints?
- If \(u_n\sim -2/n\), what is the right-endpoint verdict, and what is the sign pattern at the left endpoint?
- For \(0<p\le1\), why does \(\sum u_n\) diverge when \(u_n\sim C/n^p\) and \(C\ne0\)?
- What additional condition on \(|u_n|\) is sufficient to apply the alternating-series test at the left endpoint?
- Why does \(u_n\sim 1/n\) not, by itself, prove convergence of \(\sum(-1)^n u_n\)?
