Definition of Power Series, calculus lesson

A polynomial stops after finitely many terms. A power series keeps going, like a polynomial with infinitely many powers of $x$ . The geometric series $1 + x + x^2 + x^3 + \cdots$ is the first example. When $\lvert x \rvert < 1$ it adds up to

\frac{1}{1 - x},

but when $x = 2$ it adds up to nothing at all. The basic question about any power series is the same one. For which $x$ does it converge?

Key terms

Power series

Definition A power series is a series of the form

\sum_{n=0}^{\infty} c_n (x - a)^n = c_0 + c_1 (x - a) + c_2 (x - a)^2 + \cdots,

where $a$ and the $c_n$ are fixed real numbers and $x$ is a variable. By convention $(x - a)^0 = 1$ for every $x$ , including $x = a$ . For each number $x$ it is an ordinary series of numbers, which may converge or diverge. Where it converges, its sum defines a function of $x$ .

In plain words A power series is an infinitely long polynomial in $x - a$ . Plug in a number and you get a numerical series to test.

Example Put $x = \frac{1}{2}$ into $\sum x^n$ and you get the geometric series

1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2,

which matches $\frac{1}{1 - 1/2} = 2$ . For $x = 2$ the terms $2^n$ do not tend to $0$ , so the series diverges.

Center

Definition The number $a$ in $\sum c_n (x - a)^n$ is the center of the power series, and the series is said to be centered at $a$ , or to be a power series in $x - a$ . Every power series converges at its center, where its sum is $c_0$ , because all terms with $n \ge 1$ are $0$ .

In plain words The center is the point the series is built around. It always works there, and the question is how far from it you can go.

Example The series

\sum_{n=0}^{\infty} \frac{(x - 2)^n}{3^n}

has center $a = 2$ . At $x = 2$ it equals $1 + 0 + 0 + \cdots = 1$ .

Coefficients

Definition The numbers $c_n$ are the coefficients of the power series, and $c_n$ is the coefficient of $(x - a)^n$ . A power series is determined by its center together with its sequence of coefficients $(c_n)_{n \ge 0}$ . Missing powers have coefficient $0$ .

In plain words The coefficients are the numbers in front of the powers. To read them off, write the series so that each term is a number times $(x - a)^n$ .

Example In $\sum \frac{(x - 2)^n}{3^n}$ the coefficient is $c_n = \frac{1}{3^n}$ . In the series

\sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - \cdots

the center is $0$ , $c_{2n} = (-1)^n$ and every odd coefficient is $c_{2n+1} = 0$ .

Radius of convergence

Definition For a power series $\sum c_n (x - a)^n$ exactly one of the following holds.

It converges only at $x = a$ .
It converges absolutely for every real $x$ .
There is a number $R > 0$ such that it converges absolutely when $\lvert x - a \rvert < R$ and diverges when $\lvert x - a \rvert > R$ .

The radius of convergence is $R = 0$ in the first case, $R = \infty$ in the second and the number $R$ in the third. If all $c_n \neq 0$ and the limit

L = \lim_{n \to \infty} \left\lvert \frac{c_{n+1}}{c_n} \right\rvert

exists, finite or $\infty$ , the ratio test gives

R = \frac{1}{L},

read as $R = \infty$ when $L = 0$ and $R = 0$ when $L = \infty$ .

In plain words Around the center there is a safe zone of half width $R$ . Inside it the series converges, outside it the series diverges.

Example Take the series

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

The ratio of consecutive terms has absolute value

\frac{\lvert x - 2 \rvert^{n+1}}{(n + 1) 3^{n+1}} \cdot \frac{n \, 3^n}{\lvert x - 2 \rvert^n} = \frac{n}{n + 1} \cdot \frac{\lvert x - 2 \rvert}{3} \to \frac{\lvert x - 2 \rvert}{3}.

The ratio test gives convergence when $\lvert x - 2 \rvert < 3$ and divergence when $\lvert x - 2 \rvert > 3$ , so $R = 3$ .

Interval of convergence

Definition The interval of convergence of a power series is the set of all real $x$ at which the series converges. If $0 < R < \infty$ it is one of the four intervals

(a - R, a + R), \quad [a - R, a + R), \quad (a - R, a + R], \quad [a - R, a + R].

If $R = 0$ it is the single point $\{a\}$ , and if $R = \infty$ it is all of $\mathbb{R}$ .

In plain words The radius tells you the open middle part. Which ends belong to the interval is a separate question.

Example The series $\sum \frac{(x - 2)^n}{n \, 3^n}$ from above has $R = 3$ , so the candidates run from $-1$ to $5$ . At $x = 5$ the series is $\sum \frac{1}{n}$ , which diverges. At $x = -1$ it is $\sum \frac{(-1)^n}{n}$ , which converges by the alternating series test. The interval of convergence is $[-1, 5)$ .

Endpoint tests

Definition At an endpoint, where $\lvert x - a \rvert = R$ with $0 < R < \infty$ , the root test and the ratio test, whenever their limits exist, give the value $1$ and decide nothing. Convergence at each endpoint must be settled separately by substituting $x = a - R$ and $x = a + R$ and applying another test to the resulting numerical series, such as the divergence test, a $p$ -series or comparison test, or the alternating series test.

In plain words The ratio test is blind exactly at the edges. Plug each edge in and test that series on its own.

Example Three series with $R = 1$ and center $0$ show that every pattern happens.

Series	At $x = -1$	At $x = 1$	Interval
$\sum x^n$	terms $\pm 1$ , diverges	terms $1$ , diverges	$(-1, 1)$
$\sum \frac{x^n}{n}$	alternating, converges	harmonic, diverges	$[-1, 1)$
$\sum \frac{x^n}{n^2}$	converges absolutely	$p = 2$ , converges	$[-1, 1]$

Zero and infinite radius

Definition A power series has radius $R = 0$ when it diverges at every $x \neq a$ , and radius $R = \infty$ when it converges at every real $x$ . Both happen. When all $c_n \neq 0$ , the ratio $\lvert c_{n+1} / c_n \rvert$ tending to $\infty$ gives $R = 0$ , and the ratio tending to $0$ gives $R = \infty$ .

In plain words Some series are useless away from their center, and some work everywhere.

Example For $\sum n! \, x^n$ the ratio is $(n + 1) \lvert x \rvert$ , which tends to $\infty$ for every $x \neq 0$ , so $R = 0$ and the series converges only at $x = 0$ . For $\sum \frac{x^n}{n!}$ the ratio is $\frac{\lvert x \rvert}{n + 1} \to 0$ for every $x$ , so $R = \infty$ .

Check yourself

Find the interval of convergence of $\sum_{n=0}^{\infty} \frac{(x + 1)^n}{2^n}$ .

Show the answer

The center is $-1$ and the ratio is $\frac{\lvert x + 1 \rvert}{2}$ , so $R = 2$ . At $x = 1$ and $x = -3$ the terms are $1$ and $(-1)^n$ , which do not tend to $0$ . The interval is $(-3, 1)$ .

Find the interval of convergence of $\sum_{n=1}^{\infty} \frac{(x - 3)^n}{\sqrt{n}}$ .

Show the answer

The ratio of consecutive terms is

\sqrt{\frac{n}{n + 1}} \, \lvert x - 3 \rvert \to \lvert x - 3 \rvert,

so $R = 1$ . At $x = 4$ the series is $\sum \frac{1}{\sqrt{n}}$ , a divergent $p$ -series. At $x = 2$ it is $\sum \frac{(-1)^n}{\sqrt{n}}$ , which converges by the alternating series test. The interval is $[2, 4)$ .

What is the radius of convergence of $\sum_{n=0}^{\infty} \frac{n^2 x^n}{5^n}$ ?

Show the answer

The ratio of consecutive terms is

\frac{(n + 1)^2}{n^2} \cdot \frac{\lvert x \rvert}{5} \to \frac{\lvert x \rvert}{5},

so the series converges for $\lvert x \rvert < 5$ and diverges for $\lvert x \rvert > 5$ . That gives $R = 5$ .

Definition of Power Series

Key terms

Power series

Center

Coefficients

Radius of convergence

Interval of convergence

Endpoint tests

Zero and infinite radius

Check yourself

How this lesson connects

Builds on

Used later in