---
title: Power Series
module: Infinite Sequences and Series
moduleNumber: 9
lessonNumber: 4
order: 904
summary: >
  A power series is a polynomial of infinite degree whose convergence set is an
  interval centered at $a$, with a radius the Ratio Test finds and endpoints that
  must be tested by hand. Inside that interval the series represents a function
  that can be differentiated and integrated term by term, generating new
  representations from the geometric series.
topics: [Infinite Sequences and Series]
sources:
  - book: Stewart
    ref: "Ch. 11; §11.8 Power Series"
  - book: Stewart
    ref: "§11.9 Representations of Functions as Power Series"
---

Letting the terms of a series carry a variable turns the series into a function.
A **power series** is a series of the form

$$
\sum_{n=0}^{\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots,
$$

where the constants $c_n$ are the **coefficients**. For each fixed $x$ this is a
series of numbers that may converge or diverge; the sum, where it exists, is a
function of $x$. A power series looks like a polynomial with infinitely many
terms, and much of its behavior follows that analogy.

More generally, a **power series centered at $a$** is

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

adopting the convention $(x - a)^0 = 1$ even at $x = a$. Setting $x = a$ makes
every term past the first vanish, so a power series centered at $a$ _always_
converges at $x = a$, to $c_0$. The question is what other $x$ it converges for.

## The radius and interval of convergence

The convergence set of a power series is always an interval, centered at $a$. The [Ratio Test](/calculus/sequences-and-series/the-convergence-tests)
shows why. Applied to $\sum c_n (x-a)^n$, the ratio of successive terms carries a
factor $|x - a|$, so the test compares $|x - a|$ to a threshold.

For $\sum n! \, x^n$ the ratio is $|(n+1)! \, x^{n+1} / (n! \, x^n)| =
(n+1)|x| \to \infty$ for any $x \ne 0$, so the series converges only at $x = 0$
($R = 0$). For the geometric series $\sum x^n$ the ratio is $|x|$, giving
convergence on $(-1, 1)$ ($R = 1$). Between these extremes lies the third case, a
finite positive radius with endpoints to settle.

> **Worked example.** Find the interval of convergence of
> $\displaystyle\sum_{n=1}^{\infty} \frac{(x-3)^n}{n}$.
>
> The ratio of successive terms is
> $\left|\dfrac{(x-3)^{n+1}}{n+1}\cdot\dfrac{n}{(x-3)^n}\right|
> = \dfrac{n}{n+1}\,|x-3| \to |x-3|$, so convergence requires $|x - 3| < 1$,
> that is $2 < x < 4$; the radius is $R = 1$. Test the endpoints:
>
> - At $x = 4$: $\sum 1/n$, the divergent harmonic series.
> - At $x = 2$: $\sum (-1)^n/n$, convergent by the Alternating Series Test.
>
> The interval of convergence is $[2, 4)$: closed at the left endpoint, open at
> the right.

> **Theorem (Three cases).** For a power series $\sum c_n (x - a)^n$, exactly one
> holds:
> - it converges only at $x = a$;
> - it converges for all $x$;
> - there is a number $R > 0$ such that it converges for $|x - a| < R$ and
>   diverges for $|x - a| > R$.

> **Definition (Radius and interval of convergence).** The number $R$ is the
> **radius of convergence**, with $R = 0$ in the first case and $R = \infty$ in
> the second. The **interval of convergence** is the set of all $x$ for which the
> series converges.

$$
% caption: The convergence set is an interval of radius $R$ about the center
% $a$: convergent strictly inside, divergent strictly outside, and undecided at
% the two endpoints $a \pm R$ until each is tested by hand.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\definecolor{red}{HTML}{C0392B}
\draw[red!70, line width=2pt] (0,0) -- (2.6,0);
\draw[acc, line width=2.4pt] (2.6,0) -- (7.4,0);
\draw[red!70, line width=2pt] (7.4,0) -- (10,0);
\draw[black] (5,0.14) -- (5,-0.14);
\node[anchor=north] at (5,-0.16) {$a$};
\node[acc, anchor=south] at (5,0.18) {converges};
% endpoints
\fill[black] (2.6,0) circle (2.6pt);
\fill[black] (7.4,0) circle (2.6pt);
\node[anchor=north] at (2.6,-0.16) {$a - R$};
\node[anchor=north] at (7.4,-0.16) {$a + R$};
\node[black, anchor=south, font=\scriptsize] at (2.6,0.2) {test};
\node[black, anchor=south, font=\scriptsize] at (7.4,0.2) {test};
\node[red!80, anchor=south] at (1.1,0.18) {diverges};
\node[red!80, anchor=south] at (8.9,0.18) {diverges};
\draw[black, |-|] (2.6,-0.75) -- (5,-0.75);
\node[black, anchor=north, font=\scriptsize] at (3.8,-0.72) {$R$};
\end{tikzpicture}
$$

The Ratio (or Root) Test finds $R$ but says nothing at the endpoints $a \pm R$,
where it always returns the inconclusive $L = 1$. Each endpoint must be checked
by substituting it and applying a different test. All four outcomes occur:
$(a-R, a+R)$, $[a-R, a+R)$, $(a-R, a+R]$, or $[a-R, a+R]$.

> **Worked example.** Find the interval of convergence of
> $\displaystyle\sum_{n=0}^{\infty} \frac{(-3)^n x^n}{\sqrt{n+1}}$.
>
> The ratio of successive terms is
> $3|x|\sqrt{(n+1)/(n+2)} \to 3|x|$, so convergence requires $3|x| < 1$, i.e.
> $R = \tfrac{1}{3}$ and the open interval $(-\tfrac{1}{3}, \tfrac{1}{3})$. Test
> the endpoints:
>
> - At $x = -\tfrac{1}{3}$: the factors $(-3)^n(-\tfrac{1}{3})^n = 1$, leaving
>   $\sum 1/\sqrt{n+1}$, a divergent $p$-series ($p = \tfrac{1}{2}$).
> - At $x = \tfrac{1}{3}$: $(-3)^n(\tfrac{1}{3})^n = (-1)^n$, leaving
>   $\sum (-1)^n/\sqrt{n+1}$, convergent by the Alternating Series Test.
>
> The interval of convergence is $(-\tfrac{1}{3}, \tfrac{1}{3}]$: open on the
> left, closed on the right.

$$
% caption: All four interval types occur, differing only in which endpoints
% $a \pm R$ are included. A filled dot marks an included endpoint (the series
% converges there); a hollow dot marks an excluded one.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\foreach \y/\lab in {3/{$(\quad)$}, 2/{$[\quad)$}, 1/{$(\quad]$}, 0/{$[\quad]$}} {
  \draw[black, line width=1.6pt] (0.5,\y) -- (3.5,\y);
  \draw[black] (2,\y+0.12) -- (2,\y-0.12);
  \node[anchor=west, black!70] at (4.4,\y) {\lab};
}
% endpoint markers: filled = included, hollow = excluded
\draw[black, fill=white] (0.5,3) circle (2.6pt);
\draw[black, fill=white] (3.5,3) circle (2.6pt);
\fill[black] (0.5,2) circle (2.6pt);
\draw[black, fill=white] (3.5,2) circle (2.6pt);
\draw[black, fill=white] (0.5,1) circle (2.6pt);
\fill[black] (3.5,1) circle (2.6pt);
\fill[black] (0.5,0) circle (2.6pt);
\fill[black] (3.5,0) circle (2.6pt);
\node[anchor=north, black!70, font=\scriptsize] at (2,-0.28) {center $a$};
\end{tikzpicture}
$$

## Functions as power series

Power series represent functions in a form that can be differentiated and
integrated term by term. The starting point is the geometric series, read from
right to left as a formula for a function:

$$
\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots, \qquad |x| < 1.
$$

$$
% caption: The partial sums of $\sum x^n$ (polynomials) hug $f(x) = 1/(1-x)$
% more tightly as the degree grows, but only on the interval $(-1, 1)$; outside
% it they fly off.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (-2.2,0) -- (2.2,0) node[anchor=north east] {$x$};
\draw[black, ->] (0,-0.4) -- (0,3.9) node[anchor=north east] {$y$};
\draw[black, dashed] (1,0) -- (1,3.9);
\node[black, anchor=south west, font=\scriptsize] at (1,0) {$x = 1$};
% f(x)=1/(1-x)
\draw[black!75, very thick] plot[domain=-2.1:0.82, samples=80] (\x, {1/(1-\x)});
\node[black!75, anchor=west] at (-2.1,0.35) {$f = \tfrac{1}{1-x}$};
% partial sum s3 = 1+x+x^2+x^3
\draw[acc, thick] plot[domain=-1.35:0.95, samples=80] (\x, {1+\x+\x*\x+\x*\x*\x});
\node[acc, anchor=south east] at (0.7,3.6) {$s_3$};
% s1 = 1+x
\draw[acc!55, thick] plot[domain=-1.4:1.4, samples=2] (\x, {1+\x});
\node[acc!70, anchor=north west, font=\scriptsize] at (1.1,2.1) {$s_1$};
\end{tikzpicture}
$$

New representations come from three operations: substitution, multiplication by
a power of $x$, and factoring a constant out of the denominator.

- **Substitution.** Replacing $x$ by $-x^2$,
  $$
  \frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots, \qquad |x| < 1.
  $$
- **Factoring the denominator.** To expand $1/(2 + x)$, write it as
  $\tfrac{1}{2} \cdot \dfrac{1}{1 - (-x/2)}$ and use the geometric formula with
  ratio $-x/2$:
  $$
  \frac{1}{2 + x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}} x^n, \qquad |x| < 2.
  $$
- **Multiplying by $x^k$.** Then $x^3/(2 + x) = \sum (-1)^n x^{n+3}/2^{n+1}$ on
  the same interval, since $x^3$ can be taken inside the summation sign.

> **Worked example.** Represent $\dfrac{x}{4 + x^2}$ as a power series and give
> its interval of convergence.
>
> Factor the $4$ out of the denominator to expose a geometric ratio:
>
> $$
> \frac{1}{4 + x^2} = \frac{1}{4}\cdot\frac{1}{1 - (-x^2/4)}
> = \frac{1}{4}\sum_{n=0}^{\infty}\left(-\frac{x^2}{4}\right)^n
> = \sum_{n=0}^{\infty} \frac{(-1)^n}{4^{n+1}}\,x^{2n}.
> $$
>
> The geometric expansion is valid when $|-x^2/4| < 1$, i.e. $|x| < 2$.
> Multiplying by $x$ moves the whole series up one power:
>
> $$
> \frac{x}{4 + x^2} = \sum_{n=0}^{\infty} \frac{(-1)^n}{4^{n+1}}\,x^{2n+1},
> \qquad |x| < 2.
> $$

## Differentiating and integrating term by term

Inside its interval of convergence a power series behaves like an infinite
polynomial: it can be differentiated and integrated one term at a time, and the
radius of convergence is unchanged.

> **Theorem (Term-by-term calculus).** If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$
> has radius of convergence $R > 0$, then $f$ is differentiable on
> $(a - R, a + R)$ and
> $$
> f'(x) = \sum_{n=1}^{\infty} n\, c_n (x-a)^{n-1},
> \qquad
> \int f(x)\,\d x = C + \sum_{n=0}^{\infty} c_n \frac{(x-a)^{n+1}}{n+1}.
> $$
> Both new series have radius of convergence $R$.

The radius is preserved, but an endpoint can change status: a series may converge
at $a + R$ while its derivative diverges there. Two applications produce series
that algebraic manipulation alone does not.

**Differentiating.** Differentiating $1/(1-x) = \sum x^n$ term by term gives

$$
\frac{1}{(1 - x)^2} = \sum_{n=1}^{\infty} n\, x^{n-1} = 1 + 2x + 3x^2 + \cdots, \qquad |x| < 1.
$$

**Integrating.** Since $\dfrac{\d}{\d x}\ln(1+x) = \dfrac{1}{1+x} = 1 - x + x^2 -
\cdots$, integrating term by term (and fixing $C = 0$ from $x = 0$) gives

$$
\ln(1 + x) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}
= x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots, \qquad |x| < 1.
$$

The same route, starting from $1/(1+x^2)$, produces **Gregory's series** for the
inverse tangent:

$$
\tan^{-1} x = \int \frac{\d x}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}
= x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots, \qquad |x| \le 1.
$$

At $x = 1$ this collapses to the Leibniz formula
$\tfrac{\pi}{4} = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \cdots$.

$$
% caption: Term-by-term integration turns the series for $1/(1+x^2)$ into the
% series for $\tan^{-1} x$: each power $x^{2n}$ becomes $x^{2n+1}/(2n+1)$, with
% the radius of convergence unchanged.
\begin{tikzpicture}[scale=1.0, font=\footnotesize,
  box/.style={draw, minimum width=42mm, minimum height=11mm, align=center, font=\scriptsize}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (top) at (0,1.4) {$\dfrac{1}{1+x^2} = 1 - x^2 + x^4 - \cdots$};
\node[box] (bot) at (0,-1.4) {$\tan^{-1} x = x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$};
\draw[->, acc, thick] (top) -- (bot) node[midway, right, font=\scriptsize] {integrate each term};
\node[black, anchor=west, font=\scriptsize] at (2.5,0) {$R = 1$ both};
\end{tikzpicture}
$$

Beyond generating identities, term-by-term integration evaluates integrals with
no elementary antiderivative. Since $1/(1 + x^7) = \sum (-1)^n x^{7n}$,

$$
\int \frac{\d x}{1 + x^7} = C + \sum_{n=0}^{\infty} (-1)^n \frac{x^{7n+1}}{7n+1}
= C + x - \frac{x^8}{8} + \frac{x^{15}}{15} - \cdots,
$$

and this series, being alternating for $0 \le x < 1$, gives a definite integral
to any accuracy with a controlled error.

## Series representations obtained

Each representation above came from manipulating the geometric series, not from
computing coefficients directly. Tying the coefficients $c_n$ to the derivatives
of $f$ at the center is the route of the
[Taylor and Maclaurin series](/calculus/sequences-and-series/taylor-and-maclaurin-series).

| Function | Power series | Interval |
| --- | --- | --- |
| $\dfrac{1}{1-x}$ | $\sum x^n$ | $(-1, 1)$ |
| $\dfrac{1}{1+x^2}$ | $\sum (-1)^n x^{2n}$ | $(-1, 1)$ |
| $\dfrac{1}{(1-x)^2}$ | $\sum n\, x^{n-1}$ | $(-1, 1)$ |
| $\ln(1+x)$ | $\sum (-1)^{n-1} x^n/n$ | $(-1, 1]$ |
| $\tan^{-1} x$ | $\sum (-1)^n x^{2n+1}/(2n+1)$ | $[-1, 1]$ |