Introduction
A power series has two related but distinct features: its radius of convergence and its behavior at the boundary of its interval of convergence. The radius is determined by how quickly the coefficient magnitudes grow. Signs, however, can affect whether a series converges at a boundary point.
This distinction is useful because it lets us separate two questions. First, how far from the center does the series converge absolutely? Second, what happens at points exactly that distance from the center? We will show that replacing every coefficient by its absolute value leaves the radius unchanged, then compare positive, alternating, and irregular sign patterns.
Power Series and Their Absolute-Value Coefficients
Consider a power series centered at \(c\), with real coefficients \(a_n\):
Its coefficient absolute-value series is the power series obtained by replacing each coefficient \(a_n\) with its absolute value \(|a_n|\):
For any fixed \(x\), the magnitude of the \(n\)th term of the original series equals the magnitude of the \(n\)th term of the new series:
Thus, at every \(x\), the two series have exactly the same absolute convergence behavior. Their ordinary convergence at a boundary point need not be the same, because changing signs can affect cancellation.
Why the Radius Does Not Depend on Signs
The root-test description of the radius makes the role of signs especially clear. For the coefficients \(a_n\), define
The Cauchy–Hadamard formula gives \(R=1/L\), with the conventions \(1/0=\infty\) and \(1/\infty=0\). This formula concerns the coefficient magnitudes, not whether the coefficients are positive or negative. Here is a direct comparison of the root-test quantities for the two series.
For every \(n\), both series have \(n\)th term magnitude \(|a_n||x-c|^n\).
When \(x\ne c\), the root of this magnitude is \(|a_n|^{1/n}|x-c|\).
The limit superior is \(L|x-c|\), so the test gives absolute convergence if \(L|x-c|<1\) and divergence if \(L|x-c|>1\).
At \(x=c\), all terms with \(n\ge1\) are zero, so both series converge there, regardless of the value of \(L\).
The qualification \(x\ne c\) in Step 2 matters. At \(x=c\), every positive-degree term is zero, so its \(n\)th root is zero. If \(L=\infty\), it would be incorrect to multiply \(L\) by \(|x-c|=0\) and use that expression for the roots at the center.
Justification. At every \(x\), the absolute value of each term in either series is \(|a_n||x-c|^n\). Therefore the series of term magnitudes is identical for both. By the definition of absolute convergence, the two series converge absolutely at precisely the same points. In particular, the distances from \(c\) up to which they converge absolutely agree, so their radii are equal. Equivalently, their coefficient root quantities are identical because \(\bigl||a_n|\bigr|^{1/n}=|a_n|^{1/n}\). The center itself is included separately, since only the constant term can be nonzero there.
Worked Example: Signed Geometric Coefficients
Worked Example: Signed Geometric Coefficients
Compare the radius of \(\sum_{n=0}^{\infty}(-1)^n3^n(x-c)^n\) with that of its coefficient absolute-value series.
The absolute-value coefficients are \(3^n\). For \(x\ne c\), the \(n\)th root of the magnitude of either series’ \(n\)th term is
The root test gives absolute convergence if \(3|x-c|<1\), or \(|x-c|<\frac13\), and divergence if \(3|x-c|>1\), or \(|x-c|>\frac13\). At the center, all terms of positive degree vanish, so the series converges. Hence the radius is \(R=\frac13\) for both series.
As a check, at \(x=c+\frac16\), the root-test value is \(3\cdot\frac16=\frac12<1\), so the series converges absolutely. At \(x=c+\frac12\), it is \(3\cdot\frac12=\frac32>1\), so the terms do not tend to zero and the series diverges. The boundary values have root-test value \(1\), where the root test gives no conclusion.
Positive and Alternating Coefficients at the Boundary
A common way to see the difference between radius and endpoint behavior is to use coefficients with magnitudes \(1/n\). Consider, for \(n\ge1\), the positive-coefficient series and the alternating-coefficient series:
For either series, the \(n\)th root of the term magnitude at \(x\ne0\) is \(|x|/n^{1/n}\). Since \(n^{1/n}\to1\), its limit is \(|x|\). Both series therefore converge absolutely for \(|x|<1\) and diverge for \(|x|>1\). Their radii are both \(1\). At \(x=0\), every term is zero, so both converge there as well.
Worked Example: Positive and Alternating Boundary Behavior
Check the two boundary points \(x=1\) and \(x=-1\) for these series. At \(x=1\), the positive-coefficient series becomes \(\sum_{n=1}^{\infty}1/n\). It diverges: in each block from \(2^k\) through \(2^{k+1}-1\), there are \(2^k\) terms, each at least \(1/2^{k+1}\). The block sum is therefore at least \(1/2\), so the partial sums cannot be bounded.
At \(x=-1\), the positive-coefficient series becomes \(\sum_{n=1}^{\infty}(-1)^n/n\). It converges by the alternating series test: \(1/n\) decreases to zero. It is not absolutely convergent, since the corresponding sum of magnitudes is the divergent harmonic series.
For the alternating-coefficient series, substituting \(x=1\) gives \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\). The terms again decrease in magnitude to zero and alternate, so this converges by the alternating series test, but not absolutely. Substituting \(x=-1\) gives \(\sum_{n=1}^{\infty}(-1)^{n-1}(-1)^n/n=-\sum_{n=1}^{\infty}1/n\), which diverges. Thus, each series has radius \(1\), but their boundary behavior is reversed. The substitutions verify the signs at both endpoints explicitly.
Irregular Signs Can Also Preserve the Radius
Signs need not follow a simple repeating pattern. As long as the coefficient magnitudes remain fixed, the radius remains fixed too. The boundary sum can still respond to the sign pattern, so it is worth examining an irregular example carefully.
Worked Example: An Irregular Sign Pattern
Define coefficients \(b_n\), for \(n\ge1\), by starting with \((-1)^{n-1}/n\), then reversing its sign whenever \(n\) is a power of \(2\) greater than \(1\). The beginning is
Every coefficient has magnitude \(1/n\). Thus \(|b_n|^{1/n}=n^{-1/n}\to1\), and the series \(\sum_{n=1}^{\infty}b_nx^n\) has radius \(1\). In particular, it converges absolutely for \(|x|<1\) and diverges for \(|x|>1\).
At \(x=1\), compare its partial sums with those of the alternating harmonic series \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\), which converges by the alternating series test. At \(n=2^k\), the original alternating term is negative, and reversing it changes that term by \(2/2^k\). At all other indices, the terms are unchanged. Consequently, the difference between the two partial sums tends to
This correction is a convergent geometric series, so the partial sums of \(\sum b_n\) also converge. But the convergence is not absolute, because \(\sum |b_n|=\sum 1/n\) diverges. Therefore \(x=1\) is a point of conditional convergence. The check is that the series differs from a convergent series by a convergent correction, while its absolute-value series is exactly harmonic.
A Common Mistake: Letting Signs Determine the Radius
A tempting but incorrect claim is that the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}x^n/n\) converges for every \(x\), because its coefficients alternate. Alternation alone does not guarantee convergence for arbitrary \(x\). At \(x=2\), the magnitude of its \(n\)th term is \(2^n/n\), which does not tend to zero. Indeed, the ratio of consecutive magnitudes is
So the term magnitudes grow rather than tend to zero, and the series diverges by the term test. The correction is to find the radius from coefficient magnitudes first; signs cannot make an exterior point converge when the terms fail to tend to zero. Alternation may matter at a boundary point, where terms can tend to zero but cancellation may affect convergence.
What the Radius Tells Us—and What It Does Not
The radius is a measure of absolute convergence around the center. Changing signs without changing coefficient magnitudes cannot change this measure. But the radius alone does not settle convergence at its boundary. The positive harmonic example diverges at \(x=1\), while its alternating counterpart converges there. Both have radius \(1\).
Whenever a radius test gives a boundary value, treat that point as a separate convergence question. Substitution can change the signs of the terms; then the resulting series must be examined on its own. This is why knowing that coefficient signs do not change the radius is useful: it prevents sign patterns from distracting from the radius calculation, while leaving room for them to matter where the radius test has no verdict.
Check Your Understanding
Use the equality of term magnitudes to reason about each question, and distinguish absolute convergence from ordinary convergence.
- Why do \(\sum a_n(x-c)^n\) and \(\sum |a_n|(x-c)^n\) have the same radius?
- For \(\sum_{n=0}^{\infty}(-1)^n 5^n(x-c)^n\), what is the radius, and which root-test inequality gives it?
- Why does the root-test calculation at \(x=c\) need to be handled separately if \(L=\limsup |a_n|^{1/n}=\infty\)?
- For the positive and alternating reciprocal-coefficient series in the tutorial, which one converges at \(x=1\), and which one diverges there?
- Why can changing signs affect convergence at the boundary without changing the radius?