Radius of Convergence and Interval of Convergence Calculator
Here’s the thing about power series: knowing where they converge is half the battle. This interval of convergence calculator does the heavy lifting — enter your series, and it finds both the radius of convergence and the complete interval of convergence in seconds, with every step shown.
I built this radius of convergence calculator because computing convergence by hand is tedious and error-prone, especially when you hit endpoint tests. Whether you are working with a Taylor series centered at zero or a power series centered at some other point, the tool below handles it. Let me show you how to use it, then we will dig into the theory that makes it work.
Radius of Convergence Calculator
How to Use This Interval of Convergence Calculator
This calculator uses the ratio test (also known as d’Alembert’s ratio test) to find your radius of convergence, then tests the endpoints to determine the exact interval. Here is how to get accurate results:
Step-by-Step Instructions
- Enter the nth coefficient \( a_n \) — This is the coefficient formula WITHOUT the \( (x-a)^n \) part. Use
nas your index variable. - Set the center \( a \) — This is the point around which your power series is centered. Default is 0 (which gives you a Maclaurin series).
- Click “Calculate Convergence” — The calculator displays the radius, tests both endpoints, and gives you the complete interval with steps shown.
Input Syntax Guide
| Expression | How to Type It | Example Series |
|---|---|---|
| \( n! \) | factorial(n) | 1/factorial(n) for \( e^x \) |
| \( (-1)^n \) | (-1)^n | (-1)^n/n for \( \ln(1+x) \) |
| \( n^2 \) | n^2 | 1/n^2 for convergent p-series |
| \( 2^n \) | 2^n | n/2^n |
| Combined | (-1)^n / factorial(n) | \( \sum \frac{(-1)^n x^n}{n!} \) |
| With constants | n / 2^n | \( \sum \frac{n x^n}{2^n} \) |
Worked Example: Finding the Interval of Convergence with Steps
Let’s find the convergence of \( \displaystyle\sum_{n=1}^{\infty} \frac{n(x-3)^n}{2^n} \).
Input:
- \( a_n \):
n / 2^n - Center:
3
Step 1 — Set up the ratio test:
$$R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| = \lim_{n \to \infty} \left| \frac{n/2^n}{(n+1)/2^{n+1}} \right|$$
Step 2 — Simplify:
$$R = \lim_{n \to \infty} \frac{2n}{n+1} = 2$$
Step 3 — Determine the open interval:
The series converges when \( |x – 3| < 2 \), giving the open interval \( (1, 5) \).
Step 4 — Test the endpoints:
- At \( x = 1 \): the series becomes \( \sum (-1)^n \cdot n \), which diverges (terms do not approach zero).
- At \( x = 5 \): the series becomes \( \sum n \), which diverges.
Result: Radius of convergence \( R = 2 \). Interval of convergence: \( (1, 5) \).
What is the Radius of Convergence?
The radius of convergence is a number \( R \) that tells you exactly where a power series behaves. For any power series of the form:
$$\sum_{n=0}^{\infty} c_n (x – a)^n$$
The series will:
- Converge absolutely when \( |x – a| < R \)
- Diverge when \( |x – a| > R \)
- Require separate testing when \( |x – a| = R \) (the endpoints)
Think of \( R \) as the “safe zone” radius around the center \( a \). Inside that zone, your series converges. Outside, it diverges. On the boundary? You have to check each endpoint individually.
Three Possible Cases for the Radius
- \( R = 0 \) — The series converges only at the center point. Example: \( \sum n! \cdot x^n \).
- \( 0 < R < \infty \) — The series converges within a finite interval. This is the most common case.
- \( R = \infty \) — The series converges for all real numbers. Example: the Taylor series for \( e^x = \sum \frac{x^n}{n!} \).
Want to understand the theory behind this? See my article on D’Alembert’s Ratio Test for the mathematical foundation.
How to Find the Radius of Convergence: Two Methods
There are two standard methods for finding the radius of convergence. Both give the same answer, but one might be easier depending on the series you are working with.
Method 1: The Ratio Test (d’Alembert’s Test)
The ratio test is the workhorse method you will use most often. Given a power series with coefficients \( a_n \):
$$R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right|$$
When to use it: The ratio test works best when the coefficients involve factorials, products, or terms where consecutive ratios simplify cleanly. It is the method this calculator uses by default.
Example: For \( \sum \frac{x^n}{n!} \):
$$R = \lim_{n \to \infty} \left| \frac{1/n!}{1/(n+1)!} \right| = \lim_{n \to \infty} (n+1) = \infty$$
So the series converges for all \( x \).
Method 2: The Root Test (Cauchy-Hadamard Theorem)
The Cauchy-Hadamard theorem provides an alternative formula using the root test:
$$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}$$
When to use it: The root test is more powerful in theory (it works whenever the ratio test works, plus some cases where the ratio test fails). It is particularly useful when the coefficients involve nth powers.
Example: For \( \sum \frac{x^n}{3^n} \):
$$\frac{1}{R} = \limsup_{n \to \infty} \left(\frac{1}{3^n}\right)^{1/n} = \frac{1}{3}$$
So \( R = 3 \).
Both methods are equivalent when the limit exists. The Cauchy-Hadamard formula using \( \limsup \) is technically more general because it handles cases where the ordinary limit does not exist.
Interval of Convergence: The Complete Picture
The interval of convergence is the complete set of \( x \)-values where your series converges. It is related to the radius, but not identical — the interval includes information about what happens at the endpoints.
If your radius is \( R \) and center is \( a \), then your interval is somewhere in the range \( (a – R, a + R) \). The exact interval depends on what happens at the boundaries:
| Endpoint Behavior | Interval Notation | Example |
|---|---|---|
| Neither endpoint converges | \( (a-R, a+R) \) | \( \sum n \cdot x^n \), interval \( (-1,1) \) |
| Both endpoints converge | \( [a-R, a+R] \) | \( \sum \frac{x^n}{n^2} \), interval \( [-1,1] \) |
| Left converges, right diverges | \( [a-R, a+R) \) | \( \sum \frac{(-1)^n x^n}{n} \), interval \( (-1,1] \) |
| Right converges, left diverges | \( (a-R, a+R] \) | Less common, but possible |
How to Test Endpoints
After finding \( R \), substitute each endpoint into the original series and use convergence tests for constant series:
- Substitute \( x = a – R \) into the series. You get a series of constants.
- Substitute \( x = a + R \) into the series. You get another series of constants.
- Test each using the alternating series test, p-series test calculator, comparison test, or other applicable tests.
This is the step most students skip or get wrong. The ratio test is inconclusive at the endpoints — you must use a different test there.
Radius of Convergence for Common Series Types
Power Series
A power series is any series of the form \( \sum c_n (x – a)^n \). Every power series has a radius of convergence, and it converges absolutely inside that radius. This is the general form that includes Taylor and Maclaurin series as special cases.
Taylor Series
A Taylor series is a power series that represents a function \( f(x) \) using its derivatives at a specific point \( a \):
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x – a)^n$$
The radius of convergence of a Taylor series equals the distance from the center \( a \) to the nearest singularity of \( f(x) \) in the complex plane. You can generate Taylor series expansions using a Taylor series calculator.
Maclaurin Series
A Maclaurin series is simply a Taylor series centered at \( a = 0 \):
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n$$
The most common Maclaurin series you will encounter:
| Function | Maclaurin Series | Radius of Convergence |
|---|---|---|
| \( e^x \) | \( \sum \frac{x^n}{n!} \) | \( R = \infty \) |
| \( \sin(x) \) | \( \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!} \) | \( R = \infty \) |
| \( \cos(x) \) | \( \sum \frac{(-1)^n x^{2n}}{(2n)!} \) | \( R = \infty \) |
| \( \frac{1}{1-x} \) | \( \sum x^n \) | \( R = 1 \) |
| \( \ln(1+x) \) | \( \sum \frac{(-1)^{n+1} x^n}{n} \) | \( R = 1 \) |
| \( \arctan(x) \) | \( \sum \frac{(-1)^n x^{2n+1}}{2n+1} \) | \( R = 1 \) |
Notice that the entire functions (\( e^x \), \( \sin x \), \( \cos x \)) have infinite radius of convergence, while functions with singularities (\( \frac{1}{1-x} \) has a pole at \( x = 1 \)) have finite radius.
Step-by-Step Strategy: Finding the Interval of Convergence
Here is the complete process I recommend, which is exactly what this interval of convergence calculator with steps automates:
- Step 1: Identify the coefficients \( a_n \) and the center \( a \) of your power series.
- Step 2: Apply the ratio test or root test to find \( R \).
- Step 3: Write the open interval \( (a – R, a + R) \).
- Step 4: Substitute \( x = a – R \) into the series. Test the resulting constant series for convergence.
- Step 5: Substitute \( x = a + R \) into the series. Test the resulting constant series for convergence.
- Step 6: Write the final interval using the correct bracket notation — square brackets \( [ \) for endpoints that converge, round parentheses \( ( \) for endpoints that diverge.
If you need to test endpoints that produce p-series, use the p-series test calculator. For more complex problems, check out these math solvers that can handle symbolic computation.
Radius of Convergence in Real Analysis
Let me give you the formal definition. For real-valued power series:
Let \( \psi \in \mathbb{R} \) be a real number. Consider the power series about \( \psi \):
$$S(x) = \sum_{n=0}^{\infty} a_n (x – \psi)^n$$
Let \( I \) be the interval of convergence of \( S(x) \), with endpoints \( \psi – R \) and \( \psi + R \).
Since \( \psi \) is the midpoint of \( I \), the value \( R \) is called the radius of convergence of \( S(x) \).
A key result from real analysis (related to Abel’s theorem) is that within the open interval of convergence, the power series converges uniformly on every compact subset. This means you can differentiate and integrate the series term by term inside the interval.
Radius of Convergence in Complex Analysis
In the complex plane, the concept extends naturally. For \( \psi \in \mathbb{C} \) and \( z \in \mathbb{C} \), consider:
$$f(z) = \sum_{n=0}^{\infty} a_n (z – \psi)^n$$
The radius of convergence is the extended real number \( R \in \overline{\mathbb{R}} \) defined by:
$$R = \inf \left\{ |z – \psi| : z \in \mathbb{C}, \sum_{n=0}^{\infty} a_n (z – \psi)^n \text{ diverges} \right\}$$
In other words, \( R \) is the distance from the center to the nearest point where the series diverges. Geometrically, this creates a disk of convergence in the complex plane — the series converges for all \( z \) satisfying \( |z – \psi| < R \).
This explains why some real-valued functions have finite radius even when they look smooth everywhere on the real line. The function \( \frac{1}{1+x^2} \) is perfectly smooth for all real \( x \), but its Maclaurin series has \( R = 1 \) because the function has singularities at \( z = \pm i \) in the complex plane.
Common Power Series Reference Table
Here are the power series you’ll encounter most often, along with their intervals of convergence. Bookmark this — you will use it constantly.
| Function | Power Series | Interval |
|---|---|---|
| \( e^x \) | \( \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} \) | \( \mathbb{R} \) (all real numbers) |
| \( \sin(x) \) | \( \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \) | \( \mathbb{R} \) |
| \( \cos(x) \) | \( \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \) | \( \mathbb{R} \) |
| \( \sinh(x) \) | \( \displaystyle\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} \) | \( \mathbb{R} \) |
| \( \cosh(x) \) | \( \displaystyle\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \) | \( \mathbb{R} \) |
| \( \frac{1}{1-x} \) | \( \displaystyle\sum_{n=0}^{\infty} x^n \) | \( |x| < 1 \) |
| \( \frac{1}{1+x} \) | \( \displaystyle\sum_{n=0}^{\infty} (-1)^n x^n \) | \( |x| < 1 \) |
| \( \ln(1+x) \) | \( \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} \) | \( |x| < 1 \) |
| \( \ln(1-x) \) | \( \displaystyle -\sum_{n=1}^{\infty} \frac{x^n}{n} \) | \( |x| < 1 \) |
| \( \tan^{-1}(x) \) | \( \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \) | \( |x| < 1 \) |
| \( \ln(x) \) | \( \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n (x-1)^n}{n} \) | \( |x – 1| < 1 \) |
| \( \ln\left(\frac{1+x}{1-x}\right) \) | \( \displaystyle 2\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} \) | \( |x| < 1 \) |
Key Points to Remember
- Radius is always non-negative. You will get \( R \geq 0 \), with \( R = 0 \) meaning convergence only at the center, and \( R = \infty \) meaning convergence everywhere.
- Interval is not the same as radius. The radius tells you the distance from center; the interval tells you the actual set of converging \( x \)-values including endpoint behavior.
- Always test endpoints separately. The ratio test is inconclusive at \( |x – a| = R \). Use other tests (comparison, alternating series, p-series test, etc.) at the boundaries.
- The Cauchy-Hadamard theorem always works. When the ratio test fails (because the limit does not exist), the root test with \( \limsup \) still gives the correct radius.
- Complex analysis gives disks, not intervals. In \( \mathbb{C} \), the radius defines a disk of convergence centered at \( \psi \).
- Taylor series radius equals distance to nearest singularity. This is the quickest way to find \( R \) for known functions.
Frequently Asked Questions
How do you find the interval of convergence of a power series?
Apply the ratio test or root test to find the radius R, which gives the open interval (a-R, a+R). Then substitute each endpoint into the series and test convergence separately using the comparison test, alternating series test, or p-series test. The endpoint results determine whether you use square brackets or parentheses in the final interval notation.
What is the difference between radius and interval of convergence?
The radius of convergence R is a single non-negative number representing the distance from the center where convergence is guaranteed. The interval of convergence is the actual set of x-values where the series converges, written in interval notation like (a-R, a+R) or [a-R, a+R]. The interval accounts for endpoint behavior while the radius does not.
Can the radius of convergence be zero or infinity?
Yes to both. A radius of R = 0 means the series converges only at the center point, which happens with series like the sum of n! times x^n. A radius of R = infinity means the series converges for all real numbers, which occurs with entire functions like e^x = sum of x^n/n!. These are edge cases but they appear frequently in coursework.
What is the Cauchy-Hadamard theorem and when should I use it?
The Cauchy-Hadamard theorem states that 1/R = limsup |a_n|^(1/n) as n approaches infinity. It is a more general formula than the ratio test because it uses limsup instead of lim, so it works even when the ordinary limit does not exist. Use it when the ratio test is inconclusive or when coefficients involve nth powers rather than factorials.
Why do Taylor series for smooth functions sometimes have finite radius of convergence?
A function can be perfectly smooth on the real line but have singularities in the complex plane. The radius of convergence equals the distance from the center to the nearest complex singularity. For example, 1/(1+x^2) is smooth for all real x, but it has poles at x = plus or minus i, giving its Maclaurin series a radius of only R = 1.
