Power Series Theorems, calculus lesson

Once you know that

\frac{1}{1 - x} = 1 + x + x^2 + \cdots \qquad \text{for } \lvert x \rvert < 1,

you can build many more series from it without computing a single derivative. Substitute, multiply, differentiate, integrate. Each move is legal inside a certain interval, and the main skill is knowing which interval that is. Below, every series has positive radius and is centered at the same point $a$ , often $a = 0$ .

Key terms

Addition

Definition Suppose $f(x) = \sum a_n (x - a)^n$ for $\lvert x - a \rvert < R_1$ and $g(x) = \sum b_n (x - a)^n$ for $\lvert x - a \rvert < R_2$ . Then for every constant $k$ and all $x$ with $\lvert x - a \rvert < \min(R_1, R_2)$ ,

f(x) + g(x) = \sum_{n=0}^{\infty} (a_n + b_n)(x - a)^n, \qquad k f(x) = \sum_{n=0}^{\infty} k a_n (x - a)^n.

The radius of the sum is at least $\min(R_1, R_2)$ and can be larger.

In plain words Add the series term by term, matching powers. It works wherever both series work.

Example For $\lvert x \rvert < 1$ ,

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

The odd powers cancel, and the result agrees with $\frac{2}{1 - x^2}$ .

Multiplication

Definition With $f$ and $g$ as above, for $\lvert x - a \rvert < \min(R_1, R_2)$ the product is the Cauchy product

f(x) g(x) = \sum_{n=0}^{\infty} c_n (x - a)^n, \qquad c_n = \sum_{k=0}^{n} a_k b_{n-k}.

This is valid because both series converge absolutely inside their radii.

In plain words Multiply as you would two long polynomials and collect all the ways to make each power.

Example Square $\sum x^n$ . Every $a_k$ and $b_k$ is $1$ , so the coefficient $c_n$ counts the $n + 1$ terms $a_0 b_n, a_1 b_{n-1}, \ldots, a_n b_0$ . That gives $c_n = n + 1$ and

\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots, \qquad \lvert x \rvert < 1.

Division where the denominator is nonzero

Definition Let $f$ and $g$ be given by power series about $a$ with positive radii, and suppose $g(a) = b_0 \neq 0$ . Then there is an $r > 0$ such that $\frac{f}{g}$ equals a power series $\sum q_n (x - a)^n$ for $\lvert x - a \rvert < r$ . Its coefficients are found from $f = g \cdot \frac{f}{g}$ by matching powers,

a_n = \sum_{k=0}^{n} b_k q_{n-k}, \qquad \text{so} \qquad q_n = \frac{1}{b_0}\Bigl(a_n - \sum_{k=1}^{n} b_k q_{n-k}\Bigr).

The radius of the quotient can be smaller than both $R_1$ and $R_2$ , because $g$ may vanish somewhere. If $g(c) = 0$ and $f(c) \neq 0$ at a point $c$ inside both radii, the quotient series cannot converge beyond $\lvert c - a \rvert$ . It can stop even sooner, because zeros of $g$ at complex numbers count too. For example $\frac{1}{1 + x^2}$ has radius $1$ although $1 + x^2$ is never $0$ on the real line.

In plain words Long division of series works near the center as long as the denominator does not start at zero. Expect the series to stop working by the time you reach a zero of the denominator, and possibly earlier.

Example Divide $1$ by $1 - x - x^2$ . Here $b_0 = 1$ , $b_1 = -1$ , $b_2 = -1$ and the rest are $0$ , so $q_0 = 1$ , $q_1 = q_0 = 1$ and $q_n = q_{n-1} + q_{n-2}$ for $n \ge 2$ .

\frac{1}{1 - x - x^2} = 1 + x + 2x^2 + 3x^3 + 5x^4 + 8x^5 + \cdots.

The denominator is $0$ at $x = \frac{\sqrt{5} - 1}{2} \approx 0.618$ , and the series converges only for $\lvert x \rvert < 0.618\ldots$ , well inside the infinite radius of $1 - x - x^2$ .

Composition within convergence domains

Definition If $f(u) = \sum a_n u^n$ for $\lvert u \rvert < R$ , and $k \neq 0$ and $m \ge 1$ is an integer, then

f(k x^m) = \sum_{n=0}^{\infty} a_n k^n x^{mn} \qquad \text{whenever } \lvert k x^m \rvert < R.

More generally, if $g(x) = \sum b_n x^n$ with $b_0 = 0$ , then $f(g(x))$ equals the series obtained by substituting and collecting powers for every $x$ with $\sum \lvert b_n \rvert \lvert x \rvert^n < R$ .

In plain words You may plug a series into a series, as long as the inner values stay inside the region where the outer series works.

Example Put $u = -x^2$ into $\frac{1}{1 - u} = \sum u^n$ , valid for $\lvert u \rvert < 1$ .

\frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \qquad x^2 < 1, \text{ that is } \lvert x \rvert < 1.

Term-by-term differentiation and integration

Definition If $f(x) = \sum c_n (x - a)^n$ has radius of convergence $R > 0$ , then $f$ is differentiable, hence continuous, on $(a - R, a + R)$ , and there

f'(x) = \sum_{n=1}^{\infty} n c_n (x - a)^{n-1}, \qquad \int f(x) \, dx = C + \sum_{n=0}^{\infty} c_n \frac{(x - a)^{n+1}}{n + 1}.

Both new series have the same radius $R$ . Repeating the argument shows $f$ has derivatives of all orders on $(a - R, a + R)$ .

In plain words Inside the radius, a power series can be differentiated and integrated just like a polynomial, one term at a time.

Example Integrate $\frac{1}{1 + t} = \sum (-1)^n t^n$ from $0$ to $x$ , with $\lvert x \rvert < 1$ .

\ln(1 + x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n + 1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots, \qquad \lvert x \rvert < 1.

The constant is $0$ because both sides equal $0$ at $x = 0$ .

Endpoint care

Definition Differentiation and integration keep the radius of convergence, but not necessarily the behavior at the endpoints $x = a \pm R$ . Differentiating can lose endpoints and integrating can gain them, so each endpoint of the new series must be tested again.

In plain words The open interval stays the same. The ends have to be checked from scratch every time.

Example All three series below have $R = 1$ .

Series	Interval of convergence
$\sum_{n=1}^{\infty} \frac{x^n}{n}$	$[-1, 1)$
derivative $\sum_{n=1}^{\infty} x^{n-1}$	$(-1, 1)$
antiderivative $\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n + 1)}$	$[-1, 1]$

Abel’s theorem

Definition Let $f(x) = \sum c_n (x - a)^n$ have radius $R$ with $0 < R < \infty$ . If the series converges at the endpoint $x = a + R$ , then

\lim_{x \to (a + R)^-} f(x) = \sum_{n=0}^{\infty} c_n R^n.

The same holds at $x = a - R$ with the limit from the right. So a power series is continuous on its whole interval of convergence, endpoints included.

In plain words If the series still converges at an edge, its value there is what the function approaches as you walk up to the edge.

Example The series for $\ln(1 + x)$ converges at $x = 1$ by the alternating series test, and $\ln(1 + x)$ is continuous at $1$ . Abel’s theorem gives

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \lim_{x \to 1^-} \ln(1 + x) = \ln 2.

Check yourself

Find a power series for $\frac{1}{(1 + x)^2}$ and its interval of validity.

Show the answer

Differentiate $\frac{1}{1 + x} = \sum (-1)^n x^n$ term by term to get

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

Changing the sign gives

\frac{1}{(1 + x)^2} = 1 - 2x + 3x^2 - 4x^3 + \cdots \qquad \text{for } \lvert x \rvert < 1.

At $x = \pm 1$ the terms do not tend to $0$ .

Integrate the series for $\frac{1}{1 + x^2}$ to get a series for $\arctan x$ , then use Abel’s theorem at $x = 1$ .

Show the answer

Integrating from $0$ to $x$ gives

\arctan x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n + 1} \qquad \text{for } \lvert x \rvert < 1,

with constant $0$ since $\arctan 0 = 0$ . At $x = 1$ the series $1 - \frac{1}{3} + \frac{1}{5} - \cdots$ converges by the alternating series test, and $\arctan$ is continuous at $1$ , so the sum is $\arctan 1 = \frac{\pi}{4}$ .

Find the coefficient of $x^n$ in the Cauchy product of $\sum x^n$ and $\sum 2^n x^n$ , and say where the product equals $\frac{1}{(1 - x)(1 - 2x)}$ .

Show the answer

Here $a_k = 1$ and $b_k = 2^k$ , so the coefficient is

c_n = \sum_{k=0}^{n} 1 \cdot 2^{n-k} = 1 + 2 + 4 + \cdots + 2^n = 2^{n+1} - 1.

The two radii are $1$ and $\frac{1}{2}$ , so the product formula holds for $\lvert x \rvert < \frac{1}{2}$ .

Power Series Theorems

Key terms

Addition

Multiplication

Division where the denominator is nonzero

Composition within convergence domains

Term-by-term differentiation and integration

Endpoint care

Abel’s theorem

Check yourself

How this lesson connects

Builds on