Interval Of Convergence Calculator
Interval of Convergence Calculator
Slug: `/interval-of-convergence-calculator/`
Primary keyword: interval of convergence calculator
Meta title: Interval of Convergence Calculator with Endpoint Analysis
Meta description: Find the full interval of convergence, not just the radius. Both endpoints tested with the p-series and alternating series tests, brackets included.
Word count: approximately 1,000
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Radius against interval
Students lose marks at exactly one place in this topic, and it is not the algebra.
The radius R is a single number: how far from the centre convergence survives. The interval is the actual set of x values, written with brackets that record whether each endpoint belongs.
Getting from one to the other requires two extra tests that the ratio test cannot perform.
Why the ratio test quits at the boundary
Run the ratio test on a power series and it returns a limit L that depends on x. Convergence happens when L is under 1, which produces the open interval.
Substitute an endpoint and L becomes exactly 1. Every time. That is what defines the boundary. And the ratio test has nothing to say when L equals 1, so it hands the problem back to you.
Each endpoint therefore becomes a fresh problem: a series of constants, tested with whichever method suits its shape.
The procedure
1. Apply the ratio test to the coefficients and find R.
2. Write the open interval from c minus R to c plus R.
3. Substitute x equal to c minus R into the original series. Test it.
4. Substitute x equal to c plus R. Test that too.
5. Assemble the interval, using a square bracket where the endpoint converges and a round bracket where it does not.
Steps three and four are where the marks live.
Four outcomes, all real
| Series | R | Left endpoint | Right endpoint | Interval |
|---|---|---|---|---|
| --- | --- | --- | --- | --- |
| sum x^n | 1 | diverges | diverges | (-1, 1) |
| sum x^n / n | 1 | converges | diverges | [-1, 1) |
| sum x^n / n^2 | 1 | converges | converges | [-1, 1] |
| sum (-1)^n x^n / n | 1 | diverges | converges | (-1, 1] |
Four series, identical radius, four different intervals. No pattern connects R to the bracket type. You have to check.
Worked example
Take the sum of x to the n over n times three to the n.
Coefficients are one over n times three to the n. The ratio limit works out to one third, so R equals 3. Open interval from minus 3 to 3.
Right endpoint, x equal to 3. The three to the n in the denominator cancels the numerator exactly, leaving the sum of one over n. Harmonic, therefore divergent. Round bracket.
Left endpoint, x equal to minus 3. Now you get the sum of minus one to the n over n. Alternating, with terms decreasing to zero, so the alternating series test applies and it converges. Square bracket.
Interval: [-3, 3).
Note that convergence at the left endpoint is conditional, not absolute. Taking absolute values there returns the harmonic series again.
The rule in one sentence
Test both ends, every time.
Which test for which endpoint
After substitution you hold a constant series. Match its shape:
- A power of n in the denominator points to the p-series test. Converges when p exceeds 1.
- A factor of minus one to the n points to the alternating series test. Check the absolute terms decrease and approach zero.
- A messy rational expression points to limit comparison against the p-series it most resembles.
- Terms that do not approach zero end it immediately by the divergence test.
Notation that costs marks
Square bracket means the endpoint is included. Round bracket means excluded. Infinity always takes a round bracket, since it is not a number you can reach.
Writing [-3, 3] when the right endpoint diverges is wrong. Writing (-3, 3) when the left endpoint converges is also wrong, and it discards a genuine part of the answer.
When R is infinite the interval is all real numbers, and there are no endpoints to test. When R is zero the interval is the single point x equal to c.
Absolute or conditional
Endpoints frequently produce conditional convergence, so questions often ask you to classify it.
Inside the open interval, convergence is always absolute. At an endpoint it may be either. Test the absolute version separately: if the series of absolute values converges too, convergence is absolute; if not, it is conditional.
For the sum of x to the n over n at x equal to minus 1, the alternating series converges while its absolute version, the harmonic series, does not. Conditional.
Where this tool stops
Endpoint verdicts are reported for the forms the engine recognises. Series with coefficients defined recursively, or endpoints that need the integral test on an awkward function, are flagged rather than decided. Treat a flagged endpoint as unfinished work, not as a converging one.
Questions
Do I always have to test both endpoints?
Yes, whenever R is finite and positive. They can disagree.
Can both endpoints converge?
Yes. Sum of x to the n over n squared converges at both, giving a closed interval.
What if the series is centred somewhere other than zero?
Same method. Endpoints sit at c minus R and c plus R.
Why does the alternating series test work at endpoints so often?
Substituting a negative endpoint introduces minus one to the n, which is exactly the structure that test is built for.
Is the interval always symmetric about the centre?
The open part is. Bracket types can differ, so the final written interval may look asymmetric.