Introduction
A power series is centered at a particular number, and its terms depend on how far a point lies from that center. This creates a useful structure: if the series converges at a point away from its center, then it also converges at every point between that point and the center.
The key is not that the terms merely become small. Convergence at the given point makes its terms bounded, and moving closer to the center multiplies the \(n\)th term by a factor that shrinks geometrically with \(n\). That geometric factor is strong enough to guarantee absolute convergence.
Power Series and Convergence Points
A real power series centered at \(c\) has the form
where \(c\) and the coefficients \(a_n\) are fixed real numbers, and \(x\) is the variable. A number \(x_0\) is a convergence point if the series obtained by substituting \(x=x_0\) converges. It converges absolutely there if the series of absolute values of its terms converges.
The center \(c\) is always a convergence point: at \(x=c\), all terms with \(n\geq 1\) are zero, leaving only the finite term \(a_0\). The result below concerns what happens when convergence is known at a different point.
Why Moving Closer Creates a Geometric Bound
At a point \(x_0\ne c\), write the \(n\)th term as \(a_n(x_0-c)^n\). At a closer point \(y\), the corresponding term is \(a_n(y-c)^n\). For \(n\geq 1\), the magnitudes of these terms are related by
If \(y\) is strictly closer to \(c\) than \(x_0\) is, the fraction in parentheses is a number \(q\) with \(0\leq q<1\). Thus each term at \(y\) is the corresponding term at \(x_0\), in absolute value, multiplied by \(q^n\). As \(n\) increases, \(q^n\) shrinks geometrically.
Convergence at \(x_0\) ensures the terms there are bounded. The proof combines that boundedness with the shrinking geometric factor. For \(0\leq q<1\), the geometric series \(\sum_{n=1}^{\infty}q^n\) converges; its partial sums are \(q(1-q^N)/(1-q)\) when \(q<1\), and tend to \(q/(1-q)\). The case \(q=0\) is immediate.
Proof of the Closer-Point Result
Assume the series converges at \(x_0\ne c\). Its terms \(a_n(x_0-c)^n\) therefore tend to zero, so they are bounded: there is a finite constant \(M\geq 0\) such that \( |a_n(x_0-c)^n|\leq M \) for every \(n\geq 1\).
Take \(y\) with \(0<|y-c|<|x_0-c|\), and set \(q=|y-c|/|x_0-c|\). Then \(0<q<1\), and for each \(n\geq 1\), \( |a_n(y-c)^n|=|a_n(x_0-c)^n|q^n\leq Mq^n \).
For each positive integer \(N\), \( \sum_{n=1}^{N}|a_n(y-c)^n|\leq M\sum_{n=1}^{N}q^n \). The right-hand side is bounded above by \(Mq/(1-q)\), independently of \(N\). The partial sums on the left are nondecreasing and bounded, so they have a finite limit. Thus the series of absolute values converges.
The constant term \(a_0\) is finite, so adding it does not affect convergence. If \(y=c\), all terms with \(n\geq 1\) are zero, so the series converges there as well. This proves absolute convergence at every strictly closer point.
Finally, absolute convergence implies convergence: the partial sums of a series whose absolute values have a finite sum satisfy the Cauchy criterion, since the magnitude of any tail is at most the corresponding tail of the absolute-value series. Therefore the power series itself converges at each of these closer points.
Worked Examples
Worked Example: Moving Closer in a Geometric Power Series
Consider the power series \( \sum_{n=0}^{\infty}(x-2)^n/3^n \), centered at \(c=2\). At \(x_0=3\), the \(n\)th term is
This is a geometric series with ratio \(1/3\), so its sum is \(1/(1-1/3)=3/2\). Now take \(y=5/2\), which is closer to the center because \( |5/2-2|=1/2<1=|3-2| \). The \(n\)th term there is
This series converges absolutely, with sum \(1/(1-1/6)=6/5\). The distance ratio is \(q=(1/2)/1=1/2\); correspondingly, each term at \(y\) has magnitude \( (1/2)^n \) times the magnitude of the term at \(x_0\). As a check, substituting \(x=3\) gives the original ratio \(1/3\), and substituting \(x=5/2\) gives \(1/6\), with both ratios strictly less than \(1\).
Worked Example: A Convergent Point with Telescoping Terms
Define a power series centered at \(c=-1\) by
At \(x_0=1\), the \(n\)th term becomes \(1/[n(n+1)]\). Since \(1/[n(n+1)]=1/n-1/(n+1)\), its \(N\)th partial sum is
which tends to \(1\). Thus the series converges at \(x_0=1\). Consider the closer point \(y=0\), for which \(|0-(-1)|=1<2=|1-(-1)|\). At \(y=0\), the absolute value of the \(n\)th term is \(1/[2^n n(n+1)]\). Because \(n(n+1)\geq 1\),
The comparison series \(\sum_{n=1}^{\infty}1/2^n\) sums to \(1\), so the series at \(y=0\) converges absolutely. The check is direct: \(x_0=1\) produces the telescoping terms, while \(y=0\) produces terms bounded by a convergent geometric series. The distance ratio is \(q=1/2\), as the general result predicts.
Worked Example: A Point on the Opposite Side of the Center
Consider \( \sum_{n=0}^{\infty}x^n/3^n \), centered at \(c=0\). At \(x_0=-2\), its terms are \((-2/3)^n\). The geometric partial sum formula gives
and \((-2/3)^{N+1}\) tends to zero. Hence the series converges to \(1/(1+2/3)=3/5\). Now choose \(y=1\), on the opposite side of the center. Since \(|1|=1<2=|-2|\), this is a closer point. Its terms are \((1/3)^n\), so the series converges absolutely and has sum \(1/(1-1/3)=3/2\). The comparison factor is \(q=1/2\): \( (1/3)^n=(2/3)^n(1/2)^n \) in absolute value. Substitution confirms the ratios \(2/3\) at \(x_0=-2\) and \(1/3\) at \(y=1\), and both geometric sums are finite.
What “Interval” Means Here
The result says that the convergence set is order-convex: whenever it contains a point \(x_0\), it also contains every point between \(x_0\) and the center \(c\). If it contains points on both sides of \(c\), then it contains every point between those two points as well. For any such point, choose the known convergence point on the same side of the center and apply the closer-point result.
This is the interval property: there cannot be a gap between the center and a point where the series converges. The statement concerns the shape of the set, not whether a particular endpoint belongs to it. Convergence at a point does not, by this argument alone, determine convergence at a farther point; nor does the result decide what happens at boundary points. Those questions require additional analysis.
A Common Mistake: Confusing Small Terms with a Convergent Series
A common but incorrect argument is: “The terms at the known point tend to zero, so the series converges there; then the closer-point series converges too.” The first implication is false. A necessary condition for a series to converge is that its terms tend to zero, but that condition alone is not sufficient.
For a concrete example, the power series \( \sum_{n=1}^{\infty}x^n/n \) has terms \(1/n\) at \(x_0=1\), and \(1/n\) tends to zero. Yet the series at \(x_0=1\) is the harmonic series, which diverges. For instance, each block from \(2^k\) through \(2^{k+1}-1\) contains \(2^k\) terms, each at least \(1/2^{k+1}\), so each block has sum at least \(1/2\). The partial sums therefore grow without bound.
The correct argument starts with actual convergence at \(x_0\), not merely with terms tending to zero. Convergence gives a bound \(M\) on the terms. At a strictly closer point, the additional factor \(q^n\), with \(q<1\), turns that bound into \(Mq^n\), whose sum is finite. It is the geometric factor—not smallness by itself—that establishes convergence at the closer point.
Why the Interval Property Matters
The interval property lets us reason about a power series’ convergence set before calculating its exact extent. Once convergence is established at one noncentral point, every point between that point and the center is settled as well, and the convergence there is absolute. This structure is useful when testing points and when organizing later work on radii and intervals of convergence.
It also identifies the limits of the conclusion. The theorem does not claim that convergence extends beyond a known point, and it does not resolve the behavior at an endpoint. Keeping those distinctions clear prevents an interval-shaped conclusion from being mistaken for a complete description of the convergence set.
Check Your Understanding
Use the comparison between terms at a known point and terms at a closer point to answer these questions.
- Why does convergence at \(x_0\) imply that the terms \(a_n(x_0-c)^n\) are bounded?
- If \(|y-c|/|x_0-c|=q<1\), what factor relates the absolute values of the \(n\)th terms at \(y\) and \(x_0\)?
- Why does a bound of the form \( |a_n(y-c)^n|\leq Mq^n \) prove absolute convergence?
- What does the interval property tell you if a power series converges at one point to the right of its center?
- Does convergence at a known point determine convergence at a farther point or at a boundary point? Explain briefly.
