Power Series

Given a sequence {a_n} of real numbers, the series

$$\sum _{n=0}^{\mathrm{\infty }}{a}_n(x-x_0{)}^n=a_0+a_1(x-x_0)+a_2(x-x_0{)}^2+\dots$$

is called a power series around x₀. Thus the power series is a function of x provided it converges for some or all x. Of course, it converges for x = x₀. Whether it converges for other values of x depends on the choice of values of {a_n}.

We observe that is always possible to perform the following traslation x − x₀ = t, so that is always possible to study the following power series instead.

$$\sum _{n=0}^{\mathrm{\infty }}{a}_n{t}^n$$

Lemma 6.2.1 (Abel's Lemma). If the numeric series Σ a_nbⁿ, (b ≠ 0) converges, then also the power Σ a_n(x − x₀)ⁿ converges absolutely for every x ∈ |x − x₀| < |b|.

Proof. Since the numeric series Σ a_nbⁿ converges, the sequence of its terms tends to zero, and it is thus bounded: there exists a constant K > 0 such that |a_nbⁿ| ≤ K for every value of n. Thus it results

$$|a_n(x-x_0{)}^n|=|a_n{b}^n|\left|\frac{(x-x_0{)}^n}{b^n}\right|\le K{\left|\frac{x-x_0}{b}\right|}^n$$

The series Σ |(x − x₀)/b|ⁿ is a geometric series which converges for |(x − x₀)/b| < 1, that is for |x − x₀| < |b|. By the comparison test the series Σ |a_n(x − x₀)ⁿ| converges as well for |x − x₀| < |b|.  □

Abel's Lemma has an important consequence: if a power series converges at a point x₁ then it converges absolutely inside the interval |x| < |x₁|. If the series does not converge at x₂ then it does not converge over all the interval marked below:

Notice that for the case x = ±R; we cannot conclude anything, the series could converge at both points, at only one of them or neither that is the convergence interval could be of the form (−R, R), [−R, R], [−R, R).

Radius of Convergence

It turns out that, given any sequence {a_n}, one of the following proposition expressed by the folllowing theorem holds for its power series:

Theorem 6.2.2

1. The power series converges only for x = 0;

2. The power series converges absolutely for all x ∈ ℝ;

3. There is a unique positive number R such that the series converges if |x| < R and diverges if |x| > R.

The number R in case (iii) is called the radius of convergence of the power series. By convention, the radius of convergence is R = 0 in case (i) and R = ∞ in case (ii).

Proof. Let A := {x ∈ ℝ : Σ_{n ≥ 0} a_n xⁿ is convergent}. If this set is reduced to {0} we have case i). If A is not bounded above, then given t ∈ ℝ⁺, we may find x₁ ∈ A such that |x₁| > t, hence by Abel's lemma the power series Σ a_nxⁿ converges absolutely in the interval |x| < t. From the arbitrary of t ii) is proved.
Excluding the previous two cases we have that A is bounded and A ≠ {∅}. Let R = sup {|x|, x ∈ A} we have R > 0. The power series obviously does not converge for |x| > R, (then x ∉ A in this case). Given a h ∈ ℝ⁺, such that h < R, we may find x₁ ∈ A such that |x₁| > h, hence by Abel's lemma the power series Σ a_nxⁿ converges absolutely in the interval |x| < h. Since h is any number < R we have case iii).  □

Theorem 6.2.3 Given the power series Σ_{n ≥ 0} a_n xⁿ, if the limit

$$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}=\ell$$

exists, then the radius of convergence of the power series R is equal to

$$R=\{\begin{array}{ll}1/\ell ,& \text{if }\ell \ne 0,\mathrm{\infty }\\ +\mathrm{\infty },& \text{if }\ell =0\\ 0,& \text{if }\ell =\mathrm{\infty }.\end{array}$$

Proof. We have

$$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n{x}^n|}=\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}\cdot |x|=\ell \cdot |x|$$

From which follows that the power series converge ∀x ∈ ℝ if ℓ = 0, hence R = +∞; converges only for x = 0 if ℓ = +∞ hence R = 0; As last case if ℓ ∈ ℝ⁺ it converges for |x| = 1/ℓ, hence R = 1/ℓ.  □

A result similar to Proposition 6.2.3 involving using the Ratio test instead that the root test, which is usually simpler to apply.

Proposition 6.2.4. Given the power series Σ_{n ≥ 0} a_n xⁿ, if the limit

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\ell$$

exists, then the radius of convergence of the power series is R = 1/ℓ.

Proof. If we apply the Ratio Test to the power series, we get

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}{x}^{n+1}}{a_n{x}^n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right||x|=\ell \cdot |x|$$

By the Ratio test, the series is (absolutely) convergent if the limit value ℓ ⋅ |x| is less than 1, hence for |x| < 1/ℓ and nonconvergent if |x| > 1/ℓ, hence the radius of convergence is R = 1/ℓ.  □

The convergence of a power series can be studied using the criteria already already known to the reader for numeric series.

Example 6.2.4. For what values of x is the following power series convergent?

$$\sum _{n=0}^{\mathrm{\infty }}n!x^n$$

We use the ratio test.

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{(n+1)!x^{n+1}}{n!x^n}\right|=\underset{n\to \mathrm{\infty }}{lim}(n+1)|x|=\mathrm{\infty }$$

By the Ratio Test, the series diverges when x ≠ 0. Thus the given series converges only when x = 0.  ■

Example 6.2.5.

$$\sum _{n=1}^{\mathrm{\infty }}\frac{(x-2{)}^n}{n}$$

For which values of x is the series convergent? We apply the ratio test:

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}|\frac{(x-2{)}^{n+1}}{n+1}\cdot \frac{n}{(x-2{)}^n}|=\underset{n\to \mathrm{\infty }}{lim}\frac{1}{1+\frac{1}{n}}|x-2|=|x-2|$$

We have thus that

Example 6.2.6. An example already encountered is that of the geometric series:

$$\sum _{n=0}^{\mathrm{\infty }}{x}^n$$

which is absolutely convergent for |x| ≤ 1, having sum 1/(1−x). ■

Example 6.2.7. We prove that the following series

$$\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}$$

converges ∀x. For x = 0, the series converges with sum 1. Analogously to the previous examples using the ratio test:

$$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \mathrm{\infty }}{lim}|\frac{x^{n+1}}{(n+1)!}\cdot \frac{n!}{x^n}|=\underset{n\to \mathrm{\infty }}{lim}\frac{|x|}{n+1}=0$$

thus we have convergence ∀x ∈ ℝ.  ■

Example 6.2.8. Given the series

$$\sum _{n=0}^{\mathrm{\infty }}\frac{n-1}{n+1}{x}^n$$

has convergence ray R = 1, its terms are bounded by x = ±1. Thus the series converges absolutely for |x| ≤ 1 and does not converge for |x| ≤ 1.  ■

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