Power Series Interval of Convergence Finder

Math Algebra • Sequences and Series

Written by STEM Calculators Team Published December 19, 2025 Updated February 24, 2026

9. Power Series Interval Of Convergence Finder

Determines radius/interval of convergence for power series \(\sum c_n(x-a)^n\) using ratio/root logic (preset coefficient families + endpoint tester) and a convergence-zone graph.

Center a: Method: Coefficient family cₙ: Parameter r (for rⁿ or (1/r)ⁿ): Exponent p (for 1/nᵖ): Custom L (for radius only):

Run endpoint tester (when applicable) Show convergence zone on graph

Drag to pan • wheel/pinch to zoom • red dot is the test point

Ready

Convergence zone graph (number line)

x-axis: x • shaded: convergence region • endpoints shown open/closed

x: 0 sx: 60 convergence zone test point

Test point x₀:

Choose inputs and click Calculate.

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Theory: Power Series Interval of Convergence

A power series is an infinite “polynomial” centered at a real number a. It can converge for some x-values and diverge for others. The goal is to find the radius of convergence R and then decide whether each endpoint is included (the “interval”).

1) Standard form and the convergence picture

\[ \sum_{n=0}^{\infty} c_n\,(x-a)^n \]

There is a number \(R\in[0,\infty]\) such that the series converges absolutely when \(|x-a|<R\).
The series diverges when \(|x-a|>R\).
When \(|x-a|=R\), you must test each endpoint separately.

So the interval of convergence is usually \((a-R,\,a+R)\), possibly including one or both endpoints.

2) Radius from the ratio test

Let \(u_n=c_n(x-a)^n\). Then

\[ \left|\frac{u_{n+1}}{u_n}\right| = \left|\frac{c_{n+1}}{c_n}\right|\;|x-a|. \]

If the limit \(L=\lim\left|\frac{c_{n+1}}{c_n}\right|\) exists, the ratio test says the series converges when \(L|x-a|<1\). Therefore:

\[ L|x-a|<1 \quad\Longrightarrow\quad |x-a|<\frac{1}{L} \quad\Longrightarrow\quad R=\frac{1}{L}. \]

\[ L=0 \Rightarrow R=\infty, \qquad L=\infty \Rightarrow R=0. \]

3) Radius from the root test (limsup form)

Root test looks at:

\[ \sqrt[n]{|u_n|} = \sqrt[n]{|c_n|}\;|x-a|. \]

If \(G=\limsup \sqrt[n]{|c_n|}\), then convergence holds when \(G|x-a|<1\), so:

\[ R=\frac{1}{G}. \]

Root test is often cleaner when factorial or exponential growth appears in \(c_n\).

4) Endpoint testing

After you find a finite R, test:

Left endpoint: set \(x=a-R\) and study \(\sum c_n(-R)^n\).
Right endpoint: set \(x=a+R\) and study \(\sum c_n(R)^n\).

Common endpoint tools:

p-series: \(\sum 1/n^p\) converges only when \(p>1\).
Alternating test: \(\sum (-1)^n b_n\) converges if \(b_n\downarrow 0\).
Geometric recognition: \(\sum r^n\) converges only when \(|r|<1\).

5) Worked examples

Example A: \(\sum \frac{x^n}{n!}\)

\[ \left|\frac{c_{n+1}}{c_n}\right| = \frac{1/(n+1)!}{1/n!} = \frac{1}{n+1} \to 0. \]

\[ L=0 \Rightarrow R=\infty, \]

so it converges for all real x: interval \((-\infty,\infty)\).

Example B: \(\sum \frac{(x-2)^n}{n}\) (starts at n=1)

The radius is \(R=1\) (ratio behaves like 1). Endpoints:

\(x=1\): \(\sum \frac{(-1)^n}{n}\) converges (alternating harmonic).
\(x=3\): \(\sum \frac{1}{n}\) diverges (harmonic).

So the interval is \([1,3)\).

Frequently Asked Questions

What does the radius of convergence mean for a power series?

The radius of convergence R is the distance from the center a within which the power series converges (|x-a| < R). Outside that distance (|x-a| > R) the series diverges.

How do I get the interval of convergence from the radius?

Start with the open interval (a-R, a+R) when R is finite. Then test x = a-R and x = a+R separately to decide whether each endpoint is included.

Why do endpoints need separate testing?

At |x-a| = R, the ratio/root condition becomes borderline and does not automatically decide convergence. Each endpoint produces a different series that may converge or diverge depending on its form.

When does a power series converge for all real x?

If the computed radius is infinite (R = infinity), the series converges for every real x. This often happens when c_n decreases faster than any exponential, such as c_n = 1/n!.

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