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Interval of Convergence Calculator

Find the radius and interval of convergence of a power series, endpoints included
interval of convergence of sum from n=1 to infinity of (x - 2)^n / n
radius of convergence of sum of x^n / n!
interval of convergence of sum of n! x^n
radius of convergence of sum of 3^n x^n / sqrt(n)

Radius and Interval of Convergence

A power series centred at cc has the form

∑n=0∞an(x−c)n\sum_{n=0}^{\infty} a_n (x - c)^n

and the set of xx for which it converges is always one of three shapes: the single point x=cx = c, an interval centred at cc, or the whole real line. The half-width of that interval is the radius of convergence RR:

∣x−c∣<R⇒converges absolutely,∣x−c∣>R⇒diverges|x - c| < R \Rightarrow \text{converges absolutely}, \qquad |x - c| > R \Rightarrow \text{diverges}

The interval of convergence is that open interval together with whichever endpoints happen to converge. At ∣x−c∣=R|x - c| = R the ratio test is silent, so each endpoint has to be tested separately as an ordinary numerical series — and the two ends can genuinely disagree.

R=0R = 0 means convergence at the centre only; R=∞R = \infty means everywhere.

How to Find the Interval

The procedure

  1. Apply the ratio test to the absolute value of the general term:

L=lim⁡n→∞∣an+1(x−c)n+1an(x−c)n∣=∣x−c∣⋅lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n}\right| = |x - c| \cdot \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|

  1. Impose L<1L < 1 and solve for ∣x−c∣|x - c|. That inequality reads ∣x−c∣<R|x - c| < R, which hands you the radius directly.
  2. Write the open interval (c−R, c+R)(c - R,\ c + R).
  3. Test each endpoint by substituting it back into the original series and applying a numerical test — pp-series, alternating series, or the nth-term test.
  4. Assemble the interval, using a square bracket for each endpoint that converged.

Shortcuts

When ana_n contains an nn-th power, the root test R=1/lim⁡∣an∣nR = 1/\lim \sqrt[n]{|a_n|} is faster. A factorial in the denominator drives LL to 00 and gives R=∞R = \infty; a factorial in the numerator drives it to ∞\infty and gives R=0R = 0.

A useful sanity check: the series always converges at its own centre, where every term after the first is zero.

Common Mistakes to Avoid

  • Stopping at the radius: "R=1R = 1" is not an interval. The endpoints are where most of the marks are.
  • Assuming both endpoints behave alike: ∑(x−2)nn\sum \frac{(x-2)^n}{n} converges at one end and diverges at the other.
  • Using the ratio test at an endpoint: there L=1L = 1 and the test is inconclusive by construction. Substitute the value and pick a different test.
  • Forgetting to solve for xx: ∣x−c∣<R|x - c| < R still has to be unwrapped into c−R<x<c+Rc - R < x < c + R.
  • Mishandling a coefficient inside the power: for ∑3nxn\sum 3^n x^n the condition is 3∣x∣<13|x| < 1, so R=13R = \frac{1}{3}, not 33.
  • Ignoring the centre: the interval is centred at cc, not at 00, whenever the series is written in (x−c)(x - c).

Examples

Step 1: Ratio test: ∣(x−2)n+1n+1⋅n(x−2)n∣=∣x−2∣⋅nn+1\left|\dfrac{(x-2)^{n+1}}{n+1} \cdot \dfrac{n}{(x-2)^n}\right| = |x-2| \cdot \dfrac{n}{n+1}
Step 2: Take the limit: L=∣x−2∣⋅1=∣x−2∣L = |x-2| \cdot 1 = |x-2|, so convergence requires ∣x−2∣<1|x-2| < 1 and R=1R = 1
Step 3: Open interval: 1<x<31 < x < 3
Step 4: Endpoint x=3x = 3: the series becomes ∑1n\sum \frac{1}{n}, the harmonic series, which diverges
Step 5: Endpoint x=1x = 1: the series becomes ∑(−1)nn\sum \frac{(-1)^n}{n}, which converges by the alternating series test since 1/n1/n decreases to 00
Answer: R=1R = 1, interval [1,3)[1, 3)

Step 1: Ratio test: ∣xn+1(n+1)!⋅n!xn∣=∣x∣n+1\left|\dfrac{x^{n+1}}{(n+1)!} \cdot \dfrac{n!}{x^n}\right| = \dfrac{|x|}{n+1}
Step 2: Take the limit: L=lim⁡n→∞∣x∣n+1=0L = \lim_{n\to\infty} \dfrac{|x|}{n+1} = 0 for every real xx
Step 3: L=0<1L = 0 < 1 always, so the series converges for all xx and R=∞R = \infty
Step 4: There are no endpoints to test
Step 5: This series is the Maclaurin expansion of exe^x
Answer: R=∞R = \infty, interval (−∞,∞)(-\infty, \infty)

Step 1: Ratio test: ∣(n+1)! xn+1n! xn∣=(n+1)∣x∣\left|\dfrac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| = (n+1)|x|
Step 2: For any fixed x≠0x \neq 0, lim⁡n→∞(n+1)∣x∣=∞\lim_{n\to\infty}(n+1)|x| = \infty, so L>1L > 1 and the series diverges
Step 3: At x=0x = 0 every term beyond the first is 00, so the series trivially converges
Step 4: Hence R=0R = 0
Answer: R=0R = 0; converges only at x=0x = 0

Frequently Asked Questions

The radius R is a single number giving the half-width of the region where the series converges absolutely. The interval is the actual set of x values, which is the open interval of that width plus any endpoints that converge, so it requires two extra checks the radius alone does not.

Because at |x - c| = R the ratio test gives L = 1, which is inconclusive by definition. Substituting the endpoint turns the power series into an ordinary numerical series, which you then settle with a p-series, alternating series or comparison test.

Yes. A factorial in the numerator, as in sum n! x^n, forces R = 0, so the series converges only at its centre. A factorial in the denominator, as in sum x^n/n!, gives R = infinity and convergence on the whole real line.

The ratio test handles almost everything and is the standard choice. Switch to the root test when the coefficient is an nth power, such as (2n/(3n+1))^n, since taking the nth root cancels the exponent and the limit becomes trivial.

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