Power Series and the Radius of Convergence

What a Power Series Is

A power series centered at a is the sum over n, from zero upward, of a coefficient c sub n times the quantity x minus a raised to the n. The number a is the center, and the coefficients are fixed numbers that do not depend on x. Plugging in a specific value of x turns the power series into an ordinary numerical series, one you can throw at any convergence test you already know.

That last sentence is the key to the whole topic. A power series is not one series; it is a whole family of series indexed by x, and the question is which members of the family converge. Every power series converges at its own center, since substituting x equal to a makes every term after the first vanish. The real question is how far from the center that convergence extends.

Only Three Things Can Happen

There is a theorem that makes this topic tractable: the set of x where a power series converges is always an interval centered at a. Exactly one of three cases holds. The series may converge only at the center, in which case the radius R is zero. It may converge for every real x, in which case R is infinite. Or there is a positive finite R such that the series converges absolutely whenever the distance from x to the center is less than R and diverges whenever that distance exceeds R.

The word absolutely matters. Strictly inside the radius, convergence is absolute, which is the strong kind, and that is what licenses the term-by-term manipulations later in this guide. Strictly outside, the terms do not even approach zero, so divergence is emphatic. The only ambiguity in the entire picture lives at the two points exactly at distance R from the center.

Finding R with the Ratio Test

The standard method is to apply the ratio test to the series with x still in it, treating the absolute value of x minus a as a constant while n goes to infinity. Compute the limit of the absolute ratio of consecutive terms, which will come out as some non-negative number times the absolute value of x minus a. Set that expression less than one and solve for the distance from the center. The bound you get is R.

Work the series whose nth term, starting at n equals one, is the quantity x minus three to the n, divided by n times two to the n. The ratio of the term with index n plus one to the term with index n has absolute value equal to the absolute value of x minus three, divided by two, multiplied by n over n plus one. As n grows, the fraction n over n plus one tends to one, so the limit is the absolute value of x minus three over two. Requiring that to be less than one gives the absolute value of x minus three less than two, so the center is three and the radius of convergence is two. The series certainly converges on the open interval from one to five and diverges outside the closed interval.

The two extreme cases fall out of the same computation. For the sum of x to the n over n factorial, the ratio is the absolute value of x divided by n plus one, whose limit is zero for every x, so R is infinite. For the sum of n factorial times x to the n, the ratio is n plus one times the absolute value of x, which blows up unless x is zero, so R is zero and the series is useless as a function.

The Endpoints Must Be Checked Separately

This is the step that gets skipped and the step that gets graded. The ratio test is inconclusive exactly when its limit equals one, and that is precisely what happens at the two endpoints. So the ratio test can never tell you what occurs there. You must substitute each endpoint value of x into the original series and analyse the two resulting numerical series by hand, with whatever test fits.

Return to the example with center three and radius two. At x equal to five, the numerator becomes two to the n, which cancels the two to the n in the denominator, leaving the sum of one over n. That is the harmonic series, and it diverges. At x equal to one, the numerator becomes negative two to the n, and after the same cancellation the series is the sum of negative one to the n over n. That is the alternating harmonic series, which converges by the alternating series test, though only conditionally. So the interval of convergence is closed at one and open at five: from one inclusive to five exclusive.

That asymmetry is the whole reason for the extra work. The two endpoints genuinely can behave differently, and there is no shortcut that predicts which. An interval of convergence may be open at both ends, closed at both ends, or half open in either direction, and only substitution reveals which. When R is infinite there are no endpoints to check, and when R is zero the interval is the single point at the center.

Differentiating and Integrating Term by Term

Inside its interval of convergence, a power series behaves like an ordinary polynomial for the purposes of calculus. You may differentiate it term by term or integrate it term by term, and the resulting series represents the derivative or antiderivative of the original function. This is what makes power series so useful: it turns hard analytic operations into routine manipulations of exponents.

The radius of convergence survives these operations unchanged, but endpoint behaviour does not. Take the sum, from n equals one, of x to the n over n. Its radius is one, and checking endpoints shows it converges at negative one, by the alternating series test, and diverges at one, the harmonic series, so its interval is closed at negative one and open at one. Differentiating term by term gives the sum of x to the n minus one, which is the geometric series for one over one minus x. That has the same radius of one but diverges at both endpoints, so differentiation lost the endpoint that used to work.

Integration runs the other way and tends to gain endpoints, because dividing by n plus one makes the terms smaller. The same geometric series, whose interval is open at both ends, integrates to the sum of x to the n plus one over n plus one, which now converges at negative one. The rule to remember is simple: R never changes, endpoints always need rechecking.

A Working Procedure

Put it together into a fixed routine, and these problems become mechanical rather than mysterious.

• Identify the center a by reading the power of x minus a; a sign error here shifts the whole interval.
• Apply the ratio test to the terms with x included, and take the limit as n goes to infinity.
• Set the resulting expression less than one and solve for the absolute value of x minus a to read off R.
• If the limit is zero for all x, then R is infinite; if it is infinite for all non-zero x, then R is zero.
• Substitute each endpoint into the original series and test the two numerical series separately.
• State the interval with the correct brackets, and remember that convergence is absolute strictly inside it.

Frequently asked questions

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