AP®MathsCollege BoardCalculus BCRevision NotesUnit 10: Infinite Sequences and SeriesSeries Representations of FunctionsRadius & Interval of Convergence of Power Series

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

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 19 May 2026

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
- 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
    - 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 allows 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.

Answer:

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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