Power Series and Intervals of Convergence

A series whose input can move

A power series is an infinite polynomial centered at a chosen number $c$:

$$\sum_{n=0}^{\infty}a_n(x-c)^n.$$

For each fixed value of $x$, this becomes an ordinary numerical series. Some inputs make its terms small enough to converge; other inputs make them grow or fail to settle. The set of inputs for which the series converges is its interval of convergence. This interval is a property of the entire infinite expression, not simply the domain of one coefficient formula.

The center $c$ is special. At $x=c$, every positive power of $(x-c)$ is zero, so the series reduces to its constant term $a_0$ and converges automatically. Moving away from the center changes the effective multiplier in every term. The farther one moves, the more likely powers of $(x-c)$ will overwhelm the coefficients. The radius of convergence measures the distance from the center within which the series converges.

The ratio test finds the radius

Consider

$$\sum_{n=1}^{\infty}\frac{(x-2)^n}{n\,3^n}.$$

For a fixed $x$, take the ratio of absolute consecutive terms:

$$\left|\frac{a_{n+1}}{a_n}\right| =\left|\frac{x-2}{3}\right|\frac{n}{n+1}.$$

As $n\to\infty$, the factor $n/(n+1)$ approaches $1$. The ratio test gives absolute convergence when $|x-2|/3<1$, or $|x-2|<3$. Thus the radius is $R=3$ and the preliminary open interval is $-1<x<5$. The ratio test is especially effective because the complicated-looking $n$ factor largely cancels, leaving a condition on the distance from the center.

The radius tells only what happens strictly inside and strictly outside. Inside $|x-c|<R$, the ratio limit is less than one and the series converges absolutely. Outside $|x-c|>R$, it is greater than one and the terms cannot produce a convergent series. At $|x-c|=R$, the limit equals one, so the ratio test is inconclusive. This is why a radius is not yet a full interval of convergence.

There are two equivalent ways to record the interior condition. The inequality $|x-c|<R$ emphasizes distance from the center and is often the clearest form during ratio-test algebra. Solving it as $c-R<x<c+R$ exposes the two numerical endpoint values that must be substituted next. Keep both forms visible until the endpoint tests are finished. Converting too early can make it easy to test the wrong values when the center is not zero.

Endpoints are two separate numerical series

For the example above, substitute $x=5$ and $x=-1$ into the original series rather than into the ratio condition. At $x=5$,

$$\sum_{n=1}^{\infty}\frac{3^n}{n\,3^n} =\sum_{n=1}^{\infty}\frac1n,$$

which diverges. At $x=-1$,

$$\sum_{n=1}^{\infty}\frac{(-3)^n}{n\,3^n} =\sum_{n=1}^{\infty}\frac{(-1)^n}{n},$$

which converges by the alternating series test but not absolutely. Therefore the interval of convergence is

$$[-1,5).$$

The square bracket includes an endpoint where the substituted numerical series converges. The parenthesis excludes an endpoint where it diverges. Never infer matching endpoint behavior from the symmetric-looking radius. Coefficients and alternating signs can make the two endpoint series behave differently.

Radius can be zero or infinite

Not every power series has a finite positive radius. The series $\sum n!(x-c)^n$ diverges for every $x\ne c$ because factorial coefficients overwhelm any fixed nonzero power; its radius is $0$. At the other extreme, $\sum(x-c)^n/n!$ converges for every real $x$ because factorial denominators dominate; its radius is infinite. These cases fit the same ratio-test reasoning and prevent the radius from becoming a memorized assumption.

Within its interval, a power series behaves much like a polynomial: it can be added, multiplied, differentiated, and integrated term by term under the usual convergence guarantees. At an endpoint, those operations may change the behavior and require fresh testing. The interval belongs to a particular series representation, not automatically to every expression that happens to equal it on an interior region.

An interval of convergence should always be reported with notation that communicates its endpoint decisions. Writing only “radius $3$” leaves the reader unable to tell whether either endpoint is included. Writing only $-1<x<5$ would be wrong in the example because the left endpoint was shown to converge. The final interval is a conclusion assembled from three pieces of evidence: the center and radius from the ratio test, the left endpoint’s numerical series, and the right endpoint’s numerical series.

Retrieval check

Find the radius and interval of convergence for

$$\sum_{n=1}^{\infty}\frac{(x+1)^n}{n^2\,4^n}.$$

Use the ratio test to obtain the open interval, then substitute each endpoint into the original series and use an appropriate convergence test. State why the ratio test does not settle either endpoint. The next lesson shows how valid interior power series can be added, multiplied, differentiated, and integrated to create new functions and representations.