Absolute Convergence Inside and Divergence Outside the Radius

Why a Radius Separates Two Behaviors

A power series may converge for some values of its variable and diverge for others. The radius of convergence describes the dividing distance from its center: strictly inside that distance, the series converges absolutely; strictly outside it, the series diverges.

The key idea is a geometric bound. If a series converges at one distance from its center, its terms are bounded there. At a smaller distance, each term gains an extra factor that decreases geometrically. That factor is enough to force absolute convergence.

This argument also explains why the radius itself is different from points strictly inside or outside it. At the boundary, the geometric factor is no longer less than one, so the argument does not decide convergence. Boundary points must be tested separately.

Power Series and the Radius

Consider a power series centered at \(c\), with coefficients \(a_n\):

For a fixed real number \(x\), the distance from \(x\) to the center is \(r=|x-c|\). The series converges absolutely at \(x\) when the series of term magnitudes converges:

Absolute convergence depends on the distance \(r\), not on which side of \(c\) the point lies. To define the radius, collect all distances \(r\ge 0\) for which this absolute-value series converges. The radius \(R\) is the supremum of those distances. In this tutorial, we focus on the case \(0<R<\infty\).

The radius principle is about strict inequalities. A distance less than \(R\) guarantees absolute convergence. A distance greater than \(R\) guarantees divergence. Neither statement decides what happens when the distance equals \(R\).

The Geometric Bound

The comparison that drives the result is simple. Suppose a series converges at a distance \(s>0\). Its terms must tend to zero, and therefore they are bounded: there is a constant \(M\) such that \(|a_n|s^n\le M\) for every \(n\). If we move to a smaller distance \(r<s\), then \(q=r/s\) satisfies \(0\le q<1\), and

The comparison series \(\sum_{n=0}^{\infty}Mq^n\) is geometric and converges because \(q<1\). Thus the power series converges absolutely at the smaller distance. The shrinking factor \(q^n\) is the essential feature: it makes the terms small enough to be bounded by a convergent geometric series.

Proof of the Radius Principle

First suppose \(r=|x-c|<R\). Since \(R\) is the supremum of the distances of absolute convergence, there is some distance \(s>r\) at which the absolute-value series converges. (If no such \(s\) existed, \(r\) would be an upper bound for those distances, contradicting \(r<R\).) Set \(q=r/s\), so \(0\le q<1\).

The terms of a convergent series are bounded. Hence there is an \(M\ge 0\) such that \(|a_n|s^n\le M\) for all \(n\). Therefore

Because \(\sum_{n=0}^{\infty}Mq^n\) converges, the comparison test shows that \(\sum_{n=0}^{\infty}|a_n|r^n\) converges. This is precisely absolute convergence at \(x\).

Now suppose \(r=|x-c|>R\). To prove divergence, assume instead that the power series converges at \(x\). Its terms must then tend to zero and are bounded, so there is an \(M\ge 0\) such that \(|a_n|r^n\le M\) for every \(n\). Choose any \(s\) with \(0\le s<r\), and let \(q=s/r<1\). Then

The geometric series on the right converges. Comparison shows that the power series converges absolutely at every distance \(s<r\). Consequently, the supremum of its distances of absolute convergence must be at least \(r\), so \(R\ge r\). This contradicts the assumption \(r>R\). The assumption that the original series converges was therefore false: it diverges at \(x\).

Worked Examples

A Common Comparison Error

A frequent mistake is to reverse the ratio in the geometric bound. For the geometric power series \(\sum_{n=0}^{\infty}(x/2)^n\), the radius is \(2\). At \(x=3\), the terms are \((3/2)^n\). A mistaken calculation might replace \(3/2\) by its reciprocal \(2/3\), then conclude that the series converges because \(\sum(2/3)^n\) converges.

That comparison does not bound the terms of the given series: \((3/2)^n\) is not less than or equal to \((2/3)^n\). In fact, \((3/2)^n\) grows and does not tend to zero, so the series diverges by the term test. In the inward geometric bound, the ratio is the smaller distance divided by the larger distance. Reversing that order changes a number less than one into one greater than one and invalidates the argument.

What the Radius Does—and Does Not—Tell You

The radius organizes convergence by distance from the center. If one knows \(R\), no separate convergence test is needed at a point strictly inside or strictly outside it. But the radius does not decide convergence at a point exactly \(R\) units from the center. The examples show why: one boundary series can converge while another diverges, even when the radius principle applies to both.

This distinction matters whenever a complete description of a power series’ convergence is needed. First use the radius to settle all points with \(|x-c|<R\) or \(|x-c|>R\). Then test any points at \(|x-c|=R\) individually. The radius is a powerful dividing line, but it is not an endpoint test.