Power Series

⚡ Key Takeaways

A power series is an infinite series of the form $\sum_{n=0}^\infty a_n (x-c)^n$ , a "polynomial of infinite degree" centered at $c$ . Key result: every power series has a radius of convergence $R$ (possibly $0$ or $\infty$ ) inside which it converges absolutely and defines an infinitely differentiable function. Where it applies: representing functions, solving ODEs, complex analysis, generating functions, numerical computation. Who/when: Newton (~1665), Taylor (1715), Cauchy and Abel (1820s) gave the rigorous convergence theory.

💡 The Core Insight

A power series is the bridge between local data (the coefficients $a_n$ , which encode derivatives at one point) and a function on an interval. Convergence is governed by a single number $R$ , the radius of convergence, determined entirely by how fast $|a_n|$ grows. Inside $R$ the series behaves like an infinitely-long polynomial: you may differentiate, integrate, add, and multiply term-by-term. At the boundary $|x-c|=R$ anything can happen.

DefinitionPower Series

A power series centered at $c$ with coefficients $\{a_n\}_{n\ge 0}$ is the series

1

\quad \sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \cdots

Derivation hint: this is the definition; each term is a coefficient $a_n \in \mathbb{R}$ (or $\mathbb{C}$ ) times a power of the displacement $x-c$ from the center $c$ . Here $x$ is the variable, $c$ is the center, $a_n$ is the $n$ -th coefficient, and $n$ ranges over the non-negative integers. The series defines a function $f(x)$ on the set of $x$ where it converges, with $f(c)=a_0$ always (every other term vanishes at $x=c$ ).

DefinitionRadius and Interval of Convergence

The radius of convergence $R \in [0,\infty]$ is the number such that the series [1] converges absolutely for $|x-c|<R$ and diverges for $|x-c|>R$ . It is given by the Cauchy–Hadamard formula

2

\quad \frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n}

Derivation hint: apply the root test to $\sum |a_n(x-c)^n|$ ; the series converges when $\limsup |a_n(x-c)^n|^{1/n} = |x-c|\,\limsup|a_n|^{1/n} < 1$ . with conventions $1/0=\infty$ and $1/\infty=0$ . The interval of convergence is $(c-R, c+R)$ together with whichever endpoints converge — these must be tested separately.

TheoremCauchy–Hadamard Theorem

For the power series $\sum a_n(x-c)^n$ with $R$ defined by [2]: 1. The series converges absolutely for every $x$ with $|x-c|<R$ . 2. The series diverges for every $x$ with $|x-c|>R$ . 3. On any closed sub-interval $[c-r,c+r]$ with $r<R$ , convergence is uniform. No general claim is made for $|x-c|=R$ .

TheoremRatio Test for Radius

If the limit exists, the radius is also given by

3

\quad R = \lim_{n\to\infty} \left| \frac{a_n}{a_{n+1}} \right|

Derivation hint: apply the ratio test to the terms $u_n = a_n(x-c)^n$ ; convergence requires $\lim |u_{n+1}/u_n| = |x-c|\lim|a_{n+1}/a_n| < 1$ . This is easier to apply when ratios simplify, but [2] always works whereas [3] can fail when the limit does not exist.

ProofExistence of a Radius via Abel's Lemma

Step 1 — A convergence-at-one-point lemma. Suppose $\sum a_n(x_0-c)^n$ converges for some particular $x_0 \ne c$ . Then its terms are bounded: there is $M$ with $|a_n(x_0-c)^n| \le M$ for all $n$ , since terms of a convergent series tend to $0$ .

Step 2 — Comparison with a geometric series. For any $x$ with $|x-c| < |x_0-c|$ , set $t = \frac{|x-c|}{|x_0-c|} < 1$ . Then

4

\quad |a_n(x-c)^n| = |a_n(x_0-c)^n|\cdot t^n \le M\,t^n

Derivation hint: factor $(x-c)^n = (x_0-c)^n\,\big(\tfrac{x-c}{x_0-c}\big)^n$ and bound the first factor by $M$ using Step 1.

Step 3 — Conclusion. Since $\sum_{n} M t^n = \frac{M}{1-t} < \infty$ , the comparison test gives absolute convergence at $x$ . Thus convergence at $x_0$ forces convergence at every closer point.

Step 4 — Define $R$ . Let $R = \sup\{ |x_0-c| : \sum a_n(x_0-c)^n \text{ converges}\}$ . By Step 3, all points strictly inside radius $R$ converge; points strictly outside cannot converge (else they would force convergence at a point beyond $R$ , contradicting the sup). This $R$ coincides with [2]. $\blacksquare$

∎

TheoremTerm-by-Term Differentiation and Integration

If $f(x)=\sum_{n=0}^\infty a_n(x-c)^n$ has radius $R>0$ , then $f$ is infinitely differentiable on $(c-R,c+R)$ , and the derivative and antiderivative series

5

\quad f'(x) = \sum_{n=1}^{\infty} n\,a_n (x-c)^{n-1}, \qquad \int f\,dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x-c)^{n+1}

Derivation hint: uniform convergence on compact sub-intervals (Cauchy–Hadamard part 3) permits interchange of limit and derivative/integral; $\limsup(n|a_n|)^{1/n}=\limsup|a_n|^{1/n}$ shows the radius is unchanged. both have the same radius of convergence $R$ .

TheoremTaylor Coefficients are Forced

If $f(x)=\sum_{n=0}^\infty a_n(x-c)^n$ on $(c-R,c+R)$ , then differentiating [1] repeatedly and evaluating at $x=c$ gives

6

\quad a_n = \frac{f^{(n)}(c)}{n!}

Derivation hint: apply [5] $n$ times; all terms with power $<n$ vanish under differentiation, all with power $>n$ vanish at $x=c$ , leaving only $n!\,a_n$ . So a function has at most one power series representation at a given center — its Taylor series.

NH

Niels Henrik Abel

1802–1829

The Norwegian mathematician Abel, in his 1826 paper on the binomial series, gave the first fully rigorous treatment of power-series convergence at a time when "summing divergent series" was rampant. He famously complained that "divergent series are the invention of the devil." His Abel's theorem — that if $\sum a_n R^n$ converges then $f(x)\to \sum a_n R^n$ as $x\to R^-$ — rescued endpoint behavior and justified results like $\ln 2 = 1 - \tfrac12 + \tfrac13 - \cdots$ . Abel died of tuberculosis at 26, days before a letter arrived offering him a professorship in Berlin.

🧠 Intuition

Think of a power series as a dial of weights $a_n$ tuning how strongly each power $(x-c)^n$ contributes. Near the center $x=c$ , the low powers dominate and the series looks like its first few terms — a polynomial approximation. As you move away from $c$ , higher powers grow and the question becomes whether the coefficients shrink fast enough to tame them. The radius $R$ is exactly the tipping point: a competition between coefficient decay ( $|a_n|^{1/n}$ ) and distance growth ( $|x-c|$ ). The geometric series $\sum x^n = \frac{1}{1-x}$ is the prototype — it converges for $|x|<1$ and blows up exactly where the function it represents has a singularity, at $x=1$ . This is the deep moral: the radius of convergence is the distance to the nearest singularity (visible clearly in the complex plane).

Figure

Convergence behavior of a power series on the real line

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ExampleFinding the Interval of Convergence

Given: the series $\displaystyle \sum_{n=1}^{\infty} \frac{(x-2)^n}{n\,3^n}$ . Find $R$ and the full interval of convergence.

Step 1 — Apply the ratio test for the radius. Here $a_n = \frac{1}{n\,3^n}$ . Compute

\left|\frac{a_n}{a_{n+1}}\right| = \frac{(n+1)\,3^{n+1}}{n\,3^n} = 3\cdot\frac{n+1}{n} \to 3.

So $R = 3$ and the open interval is $(2-3,\,2+3) = (-1, 5)$ .

Step 2 — Test the left endpoint $x=-1$ . Then $(x-2)^n = (-3)^n$ , giving

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

the alternating harmonic series, which converges.

Step 3 — Test the right endpoint $x=5$ . Then $(x-2)^n = 3^n$ , giving

\sum_{n=1}^\infty \frac{3^n}{n\,3^n} = \sum_{n=1}^\infty \frac{1}{n},

the harmonic series, which diverges.

Answer:

\boxed{R = 3, \quad \text{interval of convergence } = [-1,\,5)}

ExampleSumming and Differentiating a Series

Given: Starting from the geometric series $\displaystyle \frac{1}{1-x} = \sum_{n=0}^\infty x^n$ for $|x|<1$ , evaluate $\displaystyle S = \sum_{n=1}^{\infty} \frac{n}{2^n}$ .

Step 1 — Differentiate term-by-term (valid for $|x|<1$ by [5]).

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

Step 2 — Multiply by $x$ to match the desired powers.

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

Step 3 — Substitute $x = \tfrac12$ (inside $|x|<1$ , so legal).

S = \sum_{n=1}^\infty \frac{n}{2^n} = \frac{1/2}{(1 - 1/2)^2} = \frac{1/2}{1/4} = 2.

Answer:

\boxed{\;\sum_{n=1}^{\infty} \frac{n}{2^n} = 2\;}

⚠️ Common mistake

Convergence does not equal representation. A function can be infinitely differentiable, have a Taylor series that converges everywhere, and yet that series may not equal the original function. Convergence of the Taylor series is necessary but not sufficient for $f$ to be analytic. You must check that the remainder term $R_n(x)\to 0$ , not merely that the series converges.

Endpoints are NOT determined by $R$ . The interval can be open, half-open, or closed. Always test $x=c\pm R$ separately — the ratio/root tests are silent there because they give the boundary case $L=1$ .

CounterexampleA Smooth Function Whose Taylor Series Lies

Define

f(x) = \begin{cases} e^{-1/x^2}, & x \ne 0 \\ 0, & x=0. \end{cases}

Every derivative $f^{(n)}(0) = 0$ (the exponential decay beats every polynomial). Therefore its Maclaurin series is

\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n = 0 + 0\cdot x + 0\cdot x^2 + \cdots = 0,

which converges (to $0$ ) for all $x$ — yet $f(x) \ne 0$ for any $x\ne 0$ . The series converges everywhere but represents $f$ only at the single point $x=0$ . Such functions are smooth ( $C^\infty$ ) but not analytic.

connection

Power series are the analytic backbone connecting several areas. They specialize to <a class="wiki-link" href="/wiki/taylor-series">Taylor Series</a> when coefficients come from derivatives via [6], and to <a class="wiki-link" href="/wiki/maclaurin-series">Maclaurin Series</a> when $c=0$ . The Geometric Series is the foundational example and the engine for deriving many others by substitution and term-by-term operations. In Complex Analysis, a function is holomorphic precisely when it is locally a power series, and the radius equals the distance to the nearest singularity in $\mathbb{C}$ . They generalize to <a class="wiki-link" href="/wiki/math-topgeo-laurent-series">Laurent Series</a> (allowing negative powers) and connect to <a class="wiki-link" href="/wiki/fourier-series">Fourier Series</a> as another way to expand functions in a basis. The convergence theory rests on the Ratio Test, Root Test, and <a class="wiki-link" href="/wiki/math-analysis-uniform-convergence">Uniform Convergence</a>, and the boundary behavior is governed by Abel's Theorem.

🔬 Where this appears

⏳ History

- ~1665 — Isaac Newton develops the generalized binomial series

(1+x)^\alpha = \sum \binom{\alpha}{n}x^n

and uses infinite series freely to compute areas and invert functions.

- 1668 — Nicholas Mercator publishes the series for

\ln(1+x)

.

- 1671 — James Gregory and (independently) Newton obtain the arctangent series, later used to compute

\pi

.

- 1715 — Brook Taylor publishes Methodus Incrementorum, formalizing what becomes Taylor's theorem, though without convergence considerations.

- 1742 — Colin Maclaurin popularizes the special case centered at

0

in his Treatise of Fluxions.

- 1748 — Leonhard Euler's Introductio treats series as a central object, deriving

e^{ix}=\cos x + i\sin x

and many identities — formally, with little rigor.

- 1821 — Augustin-Louis Cauchy's Cours d'analyse introduces rigorous limits and the root test, giving (with Hadamard later) the radius formula [2].

- 1826 — Niels Abel proves the binomial series rigorously and establishes Abel's theorem on endpoint convergence.

- 1892 — Jacques Hadamard's thesis cements the

\limsup

form of [2], now called the Cauchy–Hadamard theorem.

open-problem

The hardest questions concern the boundary $|x-c|=R$ and the gap between convergence and analytic representation.

Boundary behavior is delicate. A series may converge at some boundary points and diverge at others; characterizing the convergence set on the circle $|z-c|=R$ (in the complex case) connects to deep results like the Fatou and Carleson theorems on radial and tangential limits.
Divergent-yet-useful series. Asymptotic series (e.g., for $\int_0^\infty \frac{e^{-t}}{1+xt}dt$ ) diverge for every $x\ne 0$ yet give superb finite-term approximations — formalized by Borel summation and the theory of resurgence (Écalle), still active research.
Lacunary series and natural boundaries. Series like $\sum z^{2^n}$ converge on $|z|<1$ but cannot be analytically continued past the boundary circle at all (Hadamard's gap theorem) — the unit circle is a "wall." Determining when such natural boundaries occur remains subtle.
Multivariable and operator-valued series raise convergence questions (domains of convergence are no longer disks; "Reinhardt domains") that are genuinely harder and not fully classified.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also