From Radius to a Candidate Convergence Interval

From a Radius to an Interval

A radius of convergence is a distance from the center of a power series. To describe where the series is guaranteed to converge on the real line, translate that distance into an open interval around the center. The calculation is short, but it is important to keep three ideas separate: the center, the radius, and the two boundary points.

This tutorial assumes that the center \(c\) and radius \(R\) are already known. The task is not to calculate the radius. It is to express the guaranteed interior region as an interval and identify its boundary points. Identifying those points does not determine whether the series converges at either one.

What the Center and Radius Mean

A power series centered at \(c\) has the form

where the coefficients \(a_n\) are fixed and \(x\) is the variable. The center \(c\) is the point from which distances are measured. The radius of convergence \(R\) describes how far from \(c\) the series is guaranteed to converge.

The strict inequality matters. The condition \(|x-c|<R\) describes points strictly closer to the center than the radius. Rewriting the absolute-value inequality gives

Thus the open interval is \((c-R,c+R)\). Parentheses indicate that the interval contains points strictly between its two boundary points, not the boundary points themselves. This notation describes the region guaranteed by the radius result; it does not settle whether the power series also converges at either boundary.

Why the Endpoints Are Center Plus or Minus the Radius

The radius is a distance, so the two boundary points lie exactly \(R\) units to the left and right of \(c\). Moving left subtracts the radius; moving right adds it. The same conclusion follows directly from the inequality that describes convergence in the interior.

This is also a complete justification for the formula: the absolute-value condition and the double inequality are equivalent, and the double inequality describes exactly the open interval between \(c-R\) and \(c+R\). At the left boundary, \(x=c-R\), so \(|x-c|=|-R|=R\). At the right boundary, \(x=c+R\), so \(|x-c|=|R|=R\). Both points are exactly one radius from the center.

Worked Examples

Useful Checks on the Result

After calculating the interval, use a few simple checks to catch arithmetic or interpretation errors. The two endpoints should be in increasing order when \(R>0\), because

The interval’s midpoint should be the center:

Its width should be twice the radius:

These checks do not replace the endpoint calculations, but they can reveal a reversed interval or a mistaken use of the radius. In particular, the radius is the distance from the center to either boundary point. It is half the interval’s width, not the full width.

Common Mistake: Treating the Radius as the Width

Suppose the center is \(2\) and the radius is \(3\). A mistake is to make an interval of total width \(3\), placing \(1.5\) units on each side of the center. That gives the incorrect interval \((0.5,3.5)\). Its endpoints are only \(1.5\) units from \(2\), so it does not use the stated radius.

The correction is to place the full radius, \(3\), on each side:

The width is \(5-(-1)=6=2R\), and each boundary point is \(3\) units from the center. This confirms that the radius is a one-sided distance and the interval’s full width is twice that distance.

What the Open Interval Does—and Does Not—Say

For a finite positive radius, every real number strictly between \(c-R\) and \(c+R\) satisfies \(|x-c|<R\), so the power series converges there. At points strictly outside that interval, \(|x-c|>R\), so the radius result says that the series diverges.

The boundary points satisfy \(|x-c|=R\), not either of those strict inequalities. Consequently, the open interval is the guaranteed interior, while the behavior at its two boundary points remains undecided by the radius alone. Do not replace the parentheses with brackets unless a separate endpoint analysis establishes that a boundary point is included. Conversely, do not infer from the open interval notation that the series must diverge at the boundary points.

If the radius is zero or infinite, the picture has a limiting form: there is no nonempty interior interval when \(R=0\), while an infinite radius gives convergence for every real \(x\). The formula with two finite boundary points applies to the finite positive-radius case treated throughout the examples here.

Summary

Given a center \(c\) and finite positive radius \(R\), subtract the radius to obtain the left boundary and add it to obtain the right boundary. The resulting open interval is \((c-R,c+R)\). Check that the endpoints are \(R\) units from \(c\), their midpoint is \(c\), and their separation is \(2R\). This identifies the region where convergence is guaranteed without deciding what happens at either boundary.