14 Power Series

Highlights of this Chapter: we introduce the definition of a power series, and testing for convergence via ratios.

Polynomials are finite sums of multiples of powers of $x$ . The natural infinite analog is a power series, arising as the limit of a sequence of polynomials of increasing degree

Definition 14.1 (Power Series) A power series is a function defined as the limit of a sequence of polynomials $f (x) = \sum_{n \geq 0} a_{n} x^{n}$ for a sequence $a_{n}$ of real numbers. For each $x$ , this defines an infinite series; the domain of a power series is the subset $D \subset R$ of $x$ values where the series converges.

The simplest power series are polynomials themselves, which have $a_{n} = 0$ after some finite $N$ . Perhaps the second simplest power series is the one with $a_{n} = 1$ for all $n$ :

$f (x) = 1 + x + x^{2} + x^{3} + x^{4} + \dots + x^{n} + \dots$

This is none other than the geometric series in $x$ ! So, it converges whenever the common ratio $x$ satisfies $| x | < 1$ : its domain is the interval $(- 1, 1)$ .

Power series are an extremely versatile tool to reach beyond the arithmetic of polynomials, while staying close to the fundamental operations of addition/subtraction and multiplication/division. One of our main uses of them will be to provide efficient means of computing important functions (exponentials, logs, trigonometric functions, etc).

Definition 14.2 (Power Series Representation) Given a function $f (x)$ , a power series representation of $f$ is a series $s (x) = \sum a_{n} x^{n}$ such that $s (x) = f (x)$ whenever $s (x)$ converges.

14.1 Convergence

The most important thing to understand about a power series is its domain: where does it actually converge?

Definition 14.3 (Interval of Convergence) The domain of a power series, also called its interval of convergence, is the set of all $x$ for which it converges.

Proposition 14.1 If a power series converges at $u$ then it converges at all $v \in (- u, u)$ .

Proof.

Exercise 14.1 If a power series diverges at $u$ then it also diverges at all $v$ with $| v | > | u |$ .

Thus, the domain of a power series must be a set of the form $(- r, r)$ , $[- r, r]$ , $[- r, r)$ or $(- r, r]$ Since it can often be difficult to determine exactly what happens at the endpoints of the interval of convergence, where the series may converge either absolutely, conditionally, or not at all. Thus speaking of the radius (and avoiding the issue of convergence at endpoints) is often useful.

Definition 14.4 (Radius of Convergence) The radius of convergence of a power series is the largest value of $r > 0$ such that the series converges on $(- r, r)$ .

Corollary 14.1 (Absolute Convergence of Power Series) Let $f (x) = \sum a_{n} x^{n}$ be a power series with radius of convergence $r$ , and let $u \in (- r, r)$ . Then $f$ converges absolutely at $u$ .

Proof.

Thus within the radius of convergence, absolute convergence means that the terms of a power series can be re-arranged without changing the limiting value: infinite addition is commutative here. (Note that this may not apply at the endpoints of the interval of convergence.)

14.1.1 Finding the Radius of Convergence

Comparison has already taught us a lot about the convergence of series, but it can do a lot more for us. Indeed, rewriting $\sum a_{n} x^{n}$ as $\sum (a_{n}^{1 / n} x)^{n}$ suggests a natural comparison when $a_{n}^{1 / n}$ converges:

Proposition 14.2 Let $\sum a_{n} x^{n}$ be a power series, and $α = lim \sqrt[n]{| a_{n} |}$ . Then the radius of convergence is $r = 1 / α$ (where $α = \infty$ means $r = 0$ and $α = 0$ means convergence on all of $R$ ).

Proof. COMPARISON WITH GEOMETRIC SERIES

This test is clear and rather powerful - it applies to almost all series you will encounter in practice. But its not completely general as we assumed that $lim \sqrt[n]{| a_{n} |}$ converged as a hypothesis, and this may not be the case. Happily, the easy technical fix of replacing $lim$ with $lim sup$ (which does always exist) provides a completely general theorem.

Theorem 14.1 (Finding the Radius of Convergence: Cauchy-Hadamard) Let $\sum a_{n} x^{n}$ be a power series, and $α = lim sup \sqrt[n]{| a_{n} |}$ . Then the radius of convergence is $r = 1 / α$ (where $α = \infty$ means $r = 0$ and $α = 0$ means convergence on all of $R$ ).

Exercise 14.2 Generalize the proof of Proposition 14.2 to prove the above theorem, using the limsup.

This test is incredibly useful theoretically as it gives precise conditions on when a power series converges or diverges. But for specific series, its rather difficult to apply in practice: who wants to compute a sequence of $n^{t h}$ roots? So now, we seek a more practical test for convergence that is easy to apply in specific cases without worrying about total generality. And we find one in our other standardized comparison with geometric series, the ratio test!

Theorem 14.2 Let $\sum a_{n} x^{n}$ be a power series, and assume the sequence of ratios $\frac{a_{n + 1}}{a_{n}}$ converges to some $α \in R$ . Then the radius of convergence is $r = 1 / α$ (where $α = \infty$ means $r = 0$ and $α = 0$ means convergence on all of $R$ ).

Proof. COMPARISON WITH A GEOMETRIC SERIES

One weakness is that it relies on consecutive ratios and these aren’t even always defined: for example the series $\sum x^{2 n}$ has coefficients $1, 0, 1, 0, 1, 0, \dots$ . This however is easy to fix: we can apply the ratio test to the nonzero terms of a power series and also get a useful result:

Exercise 14.3 Let $a_{n}$ be the sequence of nonzero coefficients for a power series $\sum a_{n} x^{n_{k}}$ (where $n_{k}$ is an increasing sequence of powers, skipping any $n$ where the coefficient would have been zero). Assume the sequence of ratios $\frac{a_{n + 1}}{a_{n}}$ converges to some $α \in R$ . Prove the radius of convergence is $r = 1 / α$ (where $α = \infty$ means $r = 0$ and $α = 0$ means convergence on all of $R$ ).

14.2 Problems

14.2.1 Example Power Series

Power series provide us a means of describing functions via explicit formulas that we have not been able to thus far, by allowing a limiting process in their definition. For instance, we will soon see that the power series below is an exponential function.

Exercise 14.4 Show the power series $\sum \frac{x^{n}}{n!}$ converges for all $x \in R$ .

When a power series converges on a finite interval, its behavior at each endpoint may require a different argument than the ratio test (as that will give $1$ , and tell you nothing)

Example 14.1 Show the power series $\sum \frac{x^{n}}{n}$ has domain $[- 1, 1)$ .

Exercise 14.5 Show the power series $\sum \frac{x^{n}}{n^{2}}$ has domain $[- 1, 1]$ .

When the radius of convergence is $0$ , the power series converges at a single point:

Exercise 14.6 Show the power series $\sum n! x^{n}$ diverges for all $x \neq 0$ .

Exercise 14.7 Series $\sum 2^{n} x^{n}$ converges on $[- 1 / 2, 1 / 2)$ . Hint: substitution $y = 2 x$

Example 14.2 Where does $\sum 2^{n} x^{3 n}$ converge? Trickier! Need to worry about the exponents not being just $n$