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Use the ratio test on the series terms \(a_n\). Solve \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1\) to get \(|x - c| < R\), where \(R\) is the radius of convergence. Then test the two endpoints \(x = c - R\) and \(x = c + R\) separately, because the ratio test gives no answer there.
How to Use the Interval of Convergence Calculator
- Type the general term \(a_n\) using
nandx, such as(x-2)^n/(n*3^n). Usefactorial(n)for \(n!\). - Enter the starting value of \(n\) (usually 0 or 1).
- Press Solve to get the radius, the interval, and the endpoint tests.
Key Ideas
- A power series has the form \(\sum c_n (x – c)^n\). The number \(c\) is the center.
- The series converges inside an interval centered at \(c\). Half its width is the radius of convergence \(R\).
- \(R\) can be 0 (converges only at the center), a positive number, or \(\infty\) (converges for every \(x\)).
- At the endpoints, the series might converge or diverge. You must check each one.
Worked Examples
Example 1: \(\sum_{n=1}^{\infty} \frac{x^n}{n}\)
\(\left|\frac{a_{n+1}}{a_n}\right| = \frac{n}{n+1}|x| \to |x|\). So the series converges when \(|x| < 1\), and \(R = 1\).
At \(x = 1\): \(\sum \frac1n\) is the harmonic series, which diverges. At \(x = -1\): \(\sum \frac{(-1)^n}{n}\) converges by the alternating series test.
Interval: \([-1, 1)\).
Example 2: \(\sum_{n=1}^{\infty} \frac{(x-2)^n}{n\,3^n}\)
The ratio test gives \(\frac{|x – 2|}{3} < 1\), so \(|x - 2| < 3\) and \(R = 3\). At \(x = 5\) the series is \(\sum \frac1n\) (diverges). At \(x = -1\) it is \(\sum\frac{(-1)^n}{n}\) (converges).
Interval: \([-1, 5)\).
Example 3: \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\)
\(\left|\frac{a_{n+1}}{a_n}\right| = \frac{|x|}{n+1} \to 0\) for every \(x\). So \(R = \infty\), and the interval is \((-\infty, \infty)\). This series equals \(e^x\).
Example 4: \(\sum_{n=0}^{\infty} n!\,x^n\)
\(\left|\frac{a_{n+1}}{a_n}\right| = (n+1)|x| \to \infty\) unless \(x = 0\). So \(R = 0\), and the series converges only at \(x = 0\).
Endpoint Tests
| Test | Use it when the endpoint series… |
|---|---|
| Divergence test | has terms that do not go to 0 (then it diverges) |
| p-series test | looks like \(\sum \frac{1}{n^p}\): converges if \(p > 1\) |
| Alternating series test | alternates in sign and its terms shrink to 0 |
| Comparison tests | is close to a series you already know |
Common Mistakes
- Forgetting the endpoints. The ratio test never decides them.
- Losing the center. For \((x – 2)^n\), the interval is centered at 2, not 0.
- Mixing up \(R\) and the interval. \(R\) is half the length of the interval.
Learn the Math Behind the Interval of Convergence Calculator
A power series only works for some x-values. The interval of convergence tells you which ones, using the ratio test and endpoint checks.
Frequently Asked Questions
What is the difference between radius and interval of convergence?
The radius is a single number \(R\). The interval is the full set of \(x\)-values, like \([-1, 5)\), including the decision at each endpoint.
Why does the ratio test fail at the endpoints?
At the endpoints the limit equals exactly 1, and the ratio test only decides when the limit is less than or greater than 1.
Where are power series used?
They are used to approximate functions like \(e^x\), \(\sin x\), and \(\ln(1 + x)\), and to solve differential equations.
Related Guides
- Improper integral calculator (the integral test)
- Limit calculator
- How to study calculus on your own