Power Series Review | Convergence, Radius, and Operations

Two facts make the series method work, and this section establishes both: a power series may be differentiated term by term inside its interval of convergence, and its coefficients are uniquely determined by the function it represents. The first lets us substitute a series into a differential equation; the second lets us match coefficients afterward.

Quick Reference

Item	Statement
Partial sum	$s_{n} = \sum_{k = 1}^{n} u_{k}$
Convergent series	$s_{n} \to S$ as $n \to \infty$ ; $S$ is the sum
Divergent series	Any series that does not converge
Power series about $x_{0}$	$\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}$
Coefficients, center	The constants $a_{k}$ ; the point $x_{0}$
Convention	${(x - x_{0})}^{0} = 1$ always, so the series converges to $a_{0}$ at $x = x_{0}$
Radius of convergence	$ρ = 1 / L$ where $L = lim_{k \to \infty} \| \frac{a_{k + 1}}{a_{k}} \|$
Special cases	$L = 0$ gives $ρ = + \infty$ ; $L = + \infty$ gives $ρ = 0$
Term-by-term derivative	f'(x) = \displaystyle\sum_{k=1}^{\infty}ka_k\left(x-x_0\right)^{k-1}, same $ρ$
Taylor coefficients	$a_{k} = \frac{f^{(k)} (x_{0})}{k!}$
Identity theorem	If $\sum a_{k} {(x - x_{0})}^{k} = 0$ on an interval, then every $a_{k} = 0$

Review of Infinite Series

Suppose $u (n)$ is a function of the integers $n$ . Adding up the values of $u (k)$ for $k = 1, 2, \dots, n$ produces another function of $n$ ,

s (n) = u (1) + u (2) + \dots + u (n),

also defined for all integer values of $n$ . It is most convenient to alter the notation slightly and write

s_{n} = u_{1} + u_{2} + \dots + u_{n}, or more shortly s_{n} = \sum_{k = 1}^{n} u_{k} .

If $s_{n}$ tends to a limit $S$ as $n$ increases without bound, we have

lim_{n \to \infty} \sum_{k = 1}^{n} u_{k} = S,

usually written in one of the forms

\sum_{k = 1}^{\infty} u_{k} = S, u_{1} + u_{2} + u_{3} + \dots = S,

the dots denoting the indefinite continuance of the series of $u$ 's.

The meaning of these equations, expressed roughly, is that by adding more and more of the $u$ 's together we get nearer and nearer to the limit $S$ . In this case the series is called a convergent infinite series, and $S$ is called the sum of the series.

The precise meaning of convergence

If any small positive number $ϵ > 0$ is chosen, we can choose $N$ , depending on $ϵ$ , so that the sum of the first $N$ terms, or of any greater number of terms, lies between $S - ϵ$ and $S + ϵ$ . In symbols,

S - ϵ < s_{n} < S + ϵ whenever n \geq N .

The point of the definition is that no matter how tight a tolerance $ϵ$ someone demands, the partial sums eventually stay within it forever.

A series diverges when it does not converge.
If $s_{n} \to \infty$ or $s_{n} \to - \infty$ as $n$ increases without bound, we say the series $\sum_{k = 1}^{\infty} u_{k}$ is divergent, or diverges to $\infty$ or $- \infty$ as the case may be.
Notice that $\sum_{k = 1}^{\infty} u_{k}$ has no meaning in the ordinary sense of a sum, because carrying out the operation of adding infinitely many numbers is impossible. The symbol denotes a limit of partial sums, nothing more.

Power Series

A series of the form

a_{0} + a_{1} (x - x_{0}) + a_{2} {(x - x_{0})}^{2} + \dots

or more shortly

\sum_{k=0}^{\infty}a_k\left(x-x_0\right)^{k} \tag{i}

is called a power series in powers of $x - x_{0}$ , or a power series about $x_{0}$ , also called a power series centered at $x_{0}$ . The constants $a_{0}, a_{1}, a_{2}, \dots$ are the coefficients and $x_{0}$ is the center.

The symbol ${(x - x_{0})}^{0}$ is always interpreted as $1$ , even when $x = x_{0}$ . Therefore the series always converges at its center, where its sum is simply $a_{0}$ .

In particular, if $x_{0} = 0$ the power series is a series in powers of $x$ :

\sum_{k = 0}^{\infty} a_{k} x^{k} .

The $n$ th partial sum is

s_{n} (x) = a_{0} + a_{1} (x - x_{0}) + a_{2} {(x - x_{0})}^{2} + \dots + a_{n} {(x - x_{0})}^{n} = \sum_{k = 0}^{n} a_{k} {(x - x_{0})}^{k} .

Note that the upper limit is $n$ , matching the subscript, while the series itself runs to $\infty$ .

Radius of Convergence

Suppose that

L = lim_{k \to \infty} | \frac{a_{k + 1}}{a_{k}} |

exists or is $+ \infty$ . Then the power series $\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}$ has radius of convergence $ρ = 1 / L$ .

When $L = 0$ the radius of convergence is infinite, $ρ = + \infty$ , and when $L = + \infty$ we have $ρ = 0$ .

The series converges absolutely for $| x - x_{0} | < ρ$ and diverges for $| x - x_{0} | > ρ$ . At the two endpoints $| x - x_{0} | = ρ$ anything can happen, and each endpoint must be tested separately.

Find the radius of convergence of each series.

$\sum_{k = 0}^{\infty} \frac{x^{k}}{k!}$
$\sum_{k = 0}^{\infty} x^{k}$
$\sum_{k = 0}^{\infty} k! x^{k}$

Solution

Here $a_{k} = \frac{1}{k!}$ , so $L = lim_{k \to \infty} | \frac{1 / (k + 1)!}{1 / k!} | = lim_{k \to \infty} \frac{k!}{(k + 1)!} = lim_{k \to \infty} \frac{1}{k + 1} = 0 .$ Therefore $ρ = \infty$ : the series converges for every $x$ . This is the series for $e^{x}$ .
Here $a_{k} = 1$ , so $L = 1$ and $ρ = 1$ . This is the geometric series, converging to $\frac{1}{1 - x}$ for $| x | < 1$ and diverging at both endpoints.
Here $a_{k} = k!$ , so $L = lim_{k \to \infty} \frac{(k + 1)!}{k!} = lim_{k \to \infty} (k + 1) = \infty,$ giving $ρ = 0$ . The series converges only at $x = 0$ , and is useless as a representation of a function.

Operations on Power Series

The next two theorems are what make the series method legitimate.

Term-by-term differentiation. If $f (x)$ is defined by a power series

f (x) = a_{0} + a_{1} (x - x_{0}) + a_{2} {(x - x_{0})}^{2} + a_{3} {(x - x_{0})}^{3} + \dots = \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}

for $| x - x_{0} | < ρ$ , then

f'(x) = a_1 + 2a_2\left(x-x_0\right) + 3a_3\left(x-x_0\right)^{2} + \cdots = \sum_{k=1}^{\infty}ka_k\left(x-x_0\right)^{k-1}

for $| x - x_{0} | < ρ$ . That is, the power series obtained by differentiating term by term converges to the derivative of $f (x)$ on the same interval.

Two details are easy to get wrong. The sum now starts at $k = 1$ , because the constant term $a_{0}$ differentiates to zero. And the exponent drops to $k - 1$ while the factor $k$ appears out front.

Since the differentiated series has the same radius of convergence, the theorem may be applied again to get f'', and again for every higher derivative. A function represented by a power series is automatically infinitely differentiable inside the interval of convergence.

Coefficients are Taylor coefficients. If $f (x)$ is defined by a power series

f (x) = \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}

then

a_{k} = \frac{f^{(k)} (x_{0})}{k!}

with the conventions $f^{(0)} (x_{0}) = f (x_{0})$ and $0! = 1$ .

Why the coefficients are forced

Setting $x = x_{0}$ in the series kills every term but the first, giving $f (x_{0}) = a_{0}$ .

Differentiating once and then setting $x = x_{0}$ kills every term but the first of the new series, giving f'(x_0) = a_1.

Differentiating twice gives f''(x) = 2a_2 + 6a_3(x-x_0) + \cdots, so f''(x_0) = 2a_2, that is, a_2 = \dfrac{f''(x_0)}{2!}.

In general, differentiating $k$ times brings down the factor $k (k - 1) \dots 1 = k!$ from the $k$ th term while annihilating all lower terms, and setting $x = x_{0}$ annihilates all higher ones.

The identity theorem, and why it matters here. An immediate consequence is that a power series representation is unique. If

\sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k} = \sum_{k = 0}^{\infty} b_{k} {(x - x_{0})}^{k}

on an interval around $x_{0}$ , then $a_{k} = b_{k}$ for every $k$ . In particular, if a power series is identically zero on an interval, all of its coefficients are zero. This is exactly the step "equate coefficients of like powers of $x$ " used throughout the next two sections.

Shifting the Index

One technical manoeuvre appears in every series solution, so it is worth isolating. To add two series with different starting powers, reindex so that the exponents match.

Combine $\sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2}$ and $\sum_{k = 0}^{\infty} a_{k} x^{k}$ into a single series.

Solution

In the first series, let $n = k - 2$ , so $k = n + 2$ . When $k = 2$ we have $n = 0$ , and $k (k - 1)$ becomes $(n + 2) (n + 1)$ :

\sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2} = \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} .

The second series already has exponent $n$ after renaming $k$ to $n$ . Therefore

\sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2} + \sum_{k = 0}^{\infty} a_{k} x^{k} = \sum_{n = 0}^{\infty} [(n + 2) (n + 1) a_{n + 2} + a_{n}] x^{n} .

The bracket is the coefficient of $x^{n}$ , and setting it to zero for every $n$ gives the recurrence relation. This is precisely the computation carried out for y'' + y = 0 in Section 6.2.

Exercises

Find the radius of convergence of $\sum_{k = 0}^{\infty} \frac{(x - 2)^{k}}{3^{k}}$ .

Solution

Here $a_{k} = 3^{- k}$ , so

L = lim_{k \to \infty} | \frac{3^{- (k + 1)}}{3^{- k}} | = \frac{1}{3},

giving $ρ = 3$ . The series converges for $| x - 2 | < 3$ , that is, on $- 1 < x < 5$ .

Write $\sum_{k = 1}^{\infty} k a_{k} x^{k - 1}$ with the summation index shifted so the exponent is $n$ .

Solution

Let $n = k - 1$ , so $k = n + 1$ and $k = 1$ corresponds to $n = 0$ :

\sum_{k = 1}^{\infty} k a_{k} x^{k - 1} = \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} .

Show directly that if $\sum_{k = 0}^{\infty} a_{k} x^{k} = 0$ for all $x$ in an interval around $0$ , then $a_{0} = a_{1} = 0$ , and explain how the argument continues.

Solution

Setting $x = 0$ gives $a_{0} = 0$ , since all other terms vanish.

Differentiating term by term, $\sum_{k = 1}^{\infty} k a_{k} x^{k - 1} = 0$ on the same interval, and setting $x = 0$ gives $a_{1} = 0$ .

Differentiating again gives $2 a_{2} = 0$ , so $a_{2} = 0$ , and by induction $k! a_{k} = 0$ at each stage, forcing $a_{k} = 0$ for every $k$ . This is the identity theorem, and it is what licenses matching coefficients.

Frequently Asked Questions

What does the radius of convergence tell me?

It marks the boundary of the region where the series is meaningful. Inside $| x - x_{0} | < ρ$ the series converges and defines a differentiable function; outside it the series diverges and the symbol means nothing. For a series solution, $ρ$ tells you how far from $x_{0}$ your answer is valid.

Can I always differentiate a series term by term?

For a power series inside its interval of convergence, yes, and the resulting series has the same radius. This is a special property of power series; general series of functions do not permit it, and Fourier series in particular can fail badly.

Why is uniqueness of the coefficients so important?

Because the entire series method rests on the step "equate coefficients of like powers". That step is valid only if two series equal on an interval must have identical coefficients. Without uniqueness, the recurrence relations would carry no information.

What happens at the endpoints of the interval of convergence?

The ratio test says nothing there, and all behaviors occur. The series $\sum x^{k}$ diverges at both $x = \pm 1$ , while $\sum \frac{x^{k}}{k^{2}}$ converges at both. Each endpoint must be checked on its own with a separate test.

Does every infinitely differentiable function equal its Taylor series?

No. The standard counterexample is $f (x) = e^{- 1 / x^{2}}$ for $x \neq 0$ with $f (0) = 0$ : every derivative at the origin is zero, so its Taylor series is identically zero, yet the function is not. Functions that do equal their Taylor series near a point are called analytic there, and analyticity is exactly the hypothesis the existence theorem of Section 6.2 requires.

Power Series

Quick Reference

Review of Infinite Series

Power Series

Radius of Convergence

Operations on Power Series

Shifting the Index

Exercises

Frequently Asked Questions

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