Radius & Interval of Convergence of Power Series (College Board AP® Calculus BC): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 23 January 2025

Radius & interval of convergence of power series

What is a power series?

A power series is a series of the form
$\sum_{n = 0}^{\infty} a_{n} {(x - r)}^{n} = a_{0} + a_{1} (x - r) + a_{2} {(x - r)}^{2} + . . . + a_{n} {(x - r)}^{n} + . . .$
- where $\{a_{n}\}$ is a sequence of real numbers
- and $r$ is a real number
Note that $x$ is a variable, independent of the summation index numbers $n$
- This means that a power series represents a function in $x$
- The series sum will take on different values if different values of $x$ are substituted in
A Taylor or Maclaurin polynomial is a truncated power series
- I.e. it is a power series that stops at a particular value of $n$

What does it mean for a power series to converge?

Recall that a series of real numbers can either converge or diverge
A power series is not, however, a single series of real numbers
- The terms of the series change depending on what value of $x$ is substituted in
- The power series may converge for some values of $x$
  - and diverge for other values of $x$
A power series will either
- converge at a single point (i.e. for a single $x$ value)
  - Note that when $x = r$ , $\sum_{n = 0}^{\infty} a_{n} {(x - r)}^{n} = a_{0}$ , so the power series will always converge at that point
- or else it will have an interval of convergence (i.e. converge for a range of $x$ values)
  - This may be a finite interval
    - e.g. $- 3 < x \leq 3$
  - Or the interval may be infinite
    - i.e. the series converges for ALL real values of $x$

How do I find the radius of convergence for a power series?

The ratio test can be used to find the radius of convergence for a power series
- The radius of convergence is 'half the width' of the interval of convergence
For example, consider the series $\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} (\frac{x^{n}}{n \cdot 3^{n}}) = \frac{x}{3} - \frac{x^{2}}{18} + \frac{x^{3}}{81} - \frac{x^{4}}{324} + . . .$
- Note that this series starts at $n = 1$ , because there is no constant $a_{0}$ term at the start
- Apply the ratio test
  $\begin{array}{rcl} \lim_{n \to \infty} |\frac{(n + 1) th term}{n th term}| & = & \lim_{n \to \infty} |\frac{({(- 1)}^{n + 2} (\frac{x^{n + 1}}{(n + 1) \cdot 3^{n + 1}}))}{{(- 1)}^{n + 1} (\frac{x^{n}}{n \cdot 3^{n}})}| \\ = & \lim_{n \to \infty} |\frac{x^{n + 1}}{(n + 1) \cdot 3^{n + 1}} \cdot \frac{n \cdot 3^{n}}{x^{n}}| \\ = & \lim_{n \to \infty} |\frac{x}{3} \cdot \frac{n}{n + 1}| \\ = & |\frac{x}{3}| \end{array}$
  - I.e. because $\lim_{n \to \infty} |\frac{n}{n + 1}| = 1$
- The series converges if that limit is less than 1
  $\begin{array}{rcl} |\frac{x}{3}| & < & 1 \Rightarrow |x| < 3 \end{array}$
  - The radius of convergence is 3
  - Note that $|x| < 3$ is equivalent to $- 3 < x < 3$
  - The series converges for all $x$ in that interval
Or consider the series $\sum_{n = 0}^{\infty} \frac{x^{n}}{n!} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . .$
- Apply the ratio test
  $\begin{array}{rcl} \lim_{n \to \infty} |\frac{(n + 1) th term}{n th term}| & = & \lim_{n \to \infty} |\frac{(\frac{x^{n + 1}}{(n + 1)!})}{(\frac{x^{n}}{n!})}| \\ = & \lim_{n \to \infty} |\frac{x^{n + 1}}{(n + 1)!} \cdot \frac{n!}{x^{n}}| \\ = & \lim_{n \to \infty} |\frac{x}{n + 1}| \\ = & 0 \end{array}$
- Be careful with that last step
  - $x$ can theoretically take on any real number value
  - But for any particular value of $x$ , $x$ is just a finite real number
  - And the limit of a finite real number divided by $n + 1$ , as $n$ goes to infinity, is zero
- So the ratio test limit, for all values of $x$ , is 0
  - And $0 < 1$
- Therefore the power series converges for all real numbers $x$

Examiner Tips and Tricks

Be careful when using the ratio test with power series

The limit statement $n \to \infty$ only applies to places where $n$ appears in the limit expressions
$x$ is a variable that is independent of $n$ , and may be treated as a constant when evaluating the limit

How do I find the interval of convergence for a power series?

Knowing the radius of convergence will allow you to identify an open interval on which the power series converges
- E.g. $|x| < 3 \Leftrightarrow - 3 < x < 3$
But using the ratio test does not tell you what happens at the endpoints of the interval
- The series might converge at none, one or both of the endpoints
To fully determine the interval of convergence, each endpoint must be tested separately
- Note that this is not necessary, as in the second example above, if the ratio test already shows that the series converges for all values of $x$
- In that case the interval of convergence is $(- \infty, \infty)$
In the first example above, $\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} (\frac{x^{n}}{n \cdot 3^{n}})$ and the ratio test told us that the series converges for $- 3 < x < 3$
- Test the series for $x = - 3$
  - When $x = - 3$ the series becomes
    $\begin{array}{rcl} \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} (\frac{{(- 3)}^{n}}{n \cdot 3^{n}}) & = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot {(\frac{- 3}{3})}^{n} \cdot \frac{1}{n} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot {(- 1)}^{n} \cdot \frac{1}{n} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{2 n + 1} \cdot \frac{1}{n} \\ = & - \sum_{n = 1}^{\infty} \frac{1}{n} \end{array}$
    - Note that $2 n + 1$ is always an odd number, so ${(- 1)}^{2 n + 1}$ is always equal to $- 1$ for every $n = 1, 2, 3, . . .$
  - $\begin{array}{rcl} \sum_{n = 1}^{\infty} \frac{1}{n} \end{array}$ is the harmonic series, which diverges to $+ \infty$
    - So $\begin{array}{rcl} - \sum_{n = 1}^{\infty} \frac{1}{n} \end{array}$ diverges to $- \infty$
  - Therefore the power series diverges for $x = - 3$
- Test the series for $x = 3$
  - When $x = 3$ the series becomes
    $\begin{array}{rcl} \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} (\frac{{(3)}^{n}}{n \cdot 3^{n}}) & = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot {(\frac{3}{3})}^{n} \cdot \frac{1}{n} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot {(1)}^{n} \cdot \frac{1}{n} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{1}{n} \end{array}$
  - $\begin{array}{rcl} \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{1}{n} \end{array}$ is the alternating harmonic series, which converges
  - Therefore the power series converges for $x = 3$
- The interval of convergence for the series is $- 3 < x \leq 3$

Worked Example

A power series is given by $\sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{(x - 1)}^{n}}{n \cdot 2^{n}} = \frac{(x - 1)}{2} - \frac{{(x - 1)}^{2}}{8} + \frac{{(x - 1)}^{3}}{24} - \frac{{(x - 1)}^{4}}{64} + . . .$ .

Determine the radius of convergence and the interval of convergence for the series.

First apply the ratio test to determine the radius of convergence

$\begin{array}{rcl} \lim_{n \to \infty} |\frac{(n + 1) th term}{n th term}| & = & \lim_{n \to \infty} |\frac{({(- 1)}^{n + 2} \cdot \frac{{(x - 1)}^{n + 1}}{(n + 1) \cdot 2^{n + 1}})}{({(- 1)}^{n + 1} \cdot \frac{{(x - 1)}^{n}}{n \cdot 2^{n}})}| \\ = & \lim_{n \to \infty} |\frac{{(x - 1)}^{n + 1}}{(n + 1) \cdot 2^{n + 1}} \cdot \frac{n \cdot 2^{n}}{{(x - 1)}^{n}}| \\ = & \lim_{n \to \infty} |\frac{x - 1}{2} \cdot \frac{n}{n + 1}| \\ = & |\frac{x - 1}{2}| \end{array}$

That converges when the limit is less than 1

$|\frac{x - 1}{2}| < 1 \Rightarrow |x - 1| < 2$

Radius of convergence = 2

So the radius of convergence is 2, but note that here the corresponding interval is not centered on $x = 0$

$|x - 1| < 2 \Leftrightarrow - 2 < x - 1 < 2 \Leftrightarrow - 1 < x < 3$

The series converges on that open interval, but now we need to test the endpoints

When $x = - 1$ ,

$\begin{array}{rcl} \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{(x - 1)}^{n}}{n \cdot 2^{n}} & = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{((- 1) - 1)}^{n}}{n \cdot 2^{n}} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{(- 2)}^{n}}{n \cdot 2^{n}} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{(- 1)}^{n} \cdot 2^{n}}{n \cdot 2^{n}} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{2 n + 1} \cdot \frac{1}{n} \\ = & - \sum_{n = 1}^{\infty} \frac{1}{n} \end{array}$

That is the negative of the harmonic series,
and so diverges to negative infinity

So the series diverges for $x = - 1$

When $x = 3$ ,

$\begin{array}{rcl} \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{(x - 1)}^{n}}{n \cdot 2^{n}} & = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{{((3) - 1)}^{n}}{n \cdot 2^{n}} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{2^{n}}{n \cdot 2^{n}} \\ = & \sum_{n = 1}^{\infty} {(- 1)}^{n + 1} \cdot \frac{1}{n} \end{array}$

That is the alternating harmonic series,
which converges

So the series converges for $x = 3$

Combine the results to get the interval of convergence

The interval of convergence is $- 1 < x \leq 3$

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Radius & Interval of Convergence of Power Series (College Board AP® Calculus BC): Study Guide

Radius & interval of convergence of power series

What is a power series?

What does it mean for a power series to converge?

How do I find the radius of convergence for a power series?

How do I find the interval of convergence for a power series?

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules

Constant of Integration

Riemann Sums & Definite Integrals

Accumulation of Change

Riemann Sums

Trapezoidal sums

Accumulation Functions