Difficulty: Intermediate | Prerequisites: Power series definitions (Part 1 of these notes), ratio test, p-series test, alternating series test.
Once you know what a power series is, the next question is: for which values of x does it actually converge? This is the central problem of this section. Every power series has an interval of convergence and a radius of convergence, and finding them is a core exam skill in Calculus II. The ratio test does most of the heavy lifting, but the endpoints always require separate attention. If you are comfortable with the ratio test and the basic convergence tests (p-series, alternating series), you are ready for this material.
To find where a power series converges, apply the ratio test to get an open interval, then test each endpoint separately using other convergence tests (p-series, alternating series test, etc.). The result is the interval of convergence. The radius of convergence R is half the width of that interval.
Ratio test
A convergence test that computes L = lim as n approaches infinity of |a_(n+1) / a_n|. If L < 1, the series converges absolutely. If L > 1, the series diverges. If L = 1, the test is inconclusive. Think of it as comparing each term to the one before it: if the terms are shrinking fast enough (ratio below 1), the series converges.
Interval of convergence (I.O.C.)
The set of all x-values for which a given power series converges. It is always an interval of the form (a minus R, a plus R), possibly including one or both endpoints. In simple terms, it is the full answer to "where does this series work?"
Radius of convergence (R)
The number R such that the series converges for |x minus a| < R and diverges for |x minus a| > R. It is always half the width of the interval of convergence. Think of it as the "reach" of the series from its centre.
Endpoint
The two x-values at the edges of the interval of convergence: x = a minus R and x = a plus R. The ratio test cannot determine convergence at these points (it gives L = 1), so they must be checked individually.
Absolute convergence
A series converges absolutely if the series of absolute values of its terms also converges. This is a stronger condition than ordinary convergence and is useful when checking endpoints.
Conditional convergence
A series converges conditionally if it converges, but the series of absolute values diverges. The alternating series test often identifies this at an endpoint.
Given a power series, the goal is to find every x-value for which it converges. The approach has three steps:
Apply the ratio test to find the open interval where the series converges.
Identify the two endpoints of that interval.
Plug each endpoint into the original series and test convergence separately, using whichever convergence test applies (p-series, alternating series test, comparison, etc.).
For a power series sum of c_n times (x minus a)^n, compute:
L = lim as n approaches infinity of |a_(n+1) / a_n|
After simplifying, the x-dependent part separates out. You end up with something of the form:
L = |x minus a| times (some expression in n that simplifies to a number)
After taking the limit, you get L = |x minus a| / R, where R is some positive number that falls out of the algebra. Set L < 1 to find the open interval of convergence.
From L < 1, you get |x minus a| < R. Using the absolute value rule (|something| < b means minus b < something < b), this becomes:
a minus R < x < a plus R
This is the open interval (a minus R, a plus R). Inside this interval, the series converges. Outside it, the series diverges.
At x = a minus R and x = a plus R, the ratio test gives L = 1 (inconclusive). You must substitute each endpoint back into the original series and determine convergence using other tools.
Common tests at endpoints:
The series becomes a p-series (sum of 1/n^p): converges if p > 1, diverges if p is 1 or less.
The series becomes an alternating series: apply the alternating series test.
The series is a known divergent form (e.g. the harmonic series sum of 1/n): diverges.
Include the endpoint in the I.O.C. if the series converges there. Exclude it if the series diverges.
Find the interval of convergence of f(x) = sum from n = 1 to infinity of (minus 1)^n times (x minus 5)^n / (n times 2^n).
Here the centre is a = 5, and c_n = (minus 1)^n / (n times 2^n).
Applying the ratio test:
Compute |a_(n+1) / a_n|:
= |(minus 1)^(n+1) (x minus 5)^(n+1) / ((n+1) times 2^(n+1))| divided by |(minus 1)^n (x minus 5)^n / (n times 2^n)|
After cancellation: = |x minus 5| times n / (2(n+1))
Take the limit as n approaches infinity:
L = |x minus 5| times lim of n/(2(n+1)) = |x minus 5| / 2
(The limit of n/(n+1) is 1.)
Finding the open interval:
Set L < 1:
|x minus 5| / 2 < 1
|x minus 5| < 2
minus 2 < x minus 5 < 2
3 < x < 7
So the series converges for x in (3, 7). The radius of convergence is R = 2.
Checking x = 3 (left endpoint):
Substitute x = 3 into the series:
Sum from n = 1 to infinity of (minus 1)^n times (3 minus 5)^n / (n times 2^n)
= sum of (minus 1)^n times (minus 2)^n / (n times 2^n)
= sum of (minus 1)^n times (minus 1)^n times 2^n / (n times 2^n)
= sum of (minus 1)^(2n) / n = sum of 1/n
This is the harmonic series. It diverges (by the p-series test with p = 1). So x = 3 is excluded.
Checking x = 7 (right endpoint):
Substitute x = 7:
Sum from n = 1 to infinity of (minus 1)^n times (7 minus 5)^n / (n times 2^n)
= sum of (minus 1)^n times 2^n / (n times 2^n)
= sum of (minus 1)^n / n
This is the alternating harmonic series. It converges (by the alternating series test). So x = 7 is included.
Final answer:
The interval of convergence is (3, 7], that is, 3 < x and x is less than or equal to 7.
The value of R that emerges from the ratio test is called the radius of convergence. For a series centred at a:
The I.O.C. is always of the form (a minus R, a plus R), possibly including endpoints.
a is always the midpoint of the interval.
R is always the distance from a to either endpoint.
In the worked example above, a = 5 and R = 2, giving the interval centred at 5 stretching 2 units in each direction.
Note: some instructors prefer the root test for finding the I.O.C. Both tests give the same radius of convergence. The ratio test is more commonly used in this course.
Ratio test
L = lim as n approaches infinity of |a_(n+1) / a_n|. Converges if L < 1, diverges if L > 1, inconclusive if L = 1.
From ratio test to interval
Set L < 1. After simplification this yields |x minus a| < R, which expands to a minus R < x < a plus R.
Absolute value rule
|expression| < b is equivalent to minus b < expression < b. This is used every time you go from the ratio test result to the interval.
Radius of convergence
R = half the width of the interval of convergence. Equivalently, R is the number such that the series converges for |x minus a| < R and diverges for |x minus a| > R.
Finding the interval of convergence tells engineers and scientists the range of inputs over which a power series approximation is valid. In numerical computing, using a series outside its interval of convergence produces nonsense, which is why convergence checking is built into numerical libraries. In physics, the convergence radius often has a physical interpretation: for instance, in electromagnetism, it can correspond to the distance from a source beyond which a multipole expansion breaks down.
Students often assume the ratio test settles the endpoints. It does not. At the endpoints, L = 1, which is inconclusive. You must always test endpoints with a different method.
A common mistake is writing the interval of convergence as a radius. The I.O.C. is an interval (e.g. (3, 7]), not a single number. The radius R = 2 is a separate piece of information.
Students sometimes forget that positive factors can be pulled in and out of absolute values freely, but factors whose sign depends on x (like x minus 5) must stay inside the absolute value until the sign is resolved.
Another error is mixing up "converges" and "converges absolutely." A series can converge at an endpoint without converging absolutely (conditional convergence). Include that endpoint either way.
⚠️ "Find the interval of convergence" is one of the most common exam questions in Calculus II. Expect it.
⚠️ Show all three stages in your working: ratio test, solve the inequality, then check endpoints separately. Marks are typically awarded at each stage.
⚠️ At each endpoint, name the test you are using (alternating series test, p-series test, etc.). Exam graders look for this.
⚠️ State the final I.O.C. with correct bracket notation: square bracket for an included endpoint, round bracket for an excluded one. Getting the brackets wrong after doing all the work is a common way to lose marks.
⚠️ The centre a is always the midpoint of the I.O.C. If you get an interval that is not symmetric about the centre, you have made an algebra error.
True or false: The ratio test can determine whether a power series converges at the endpoints of its interval. (False. The ratio test gives L = 1 at endpoints, which is inconclusive.)
Fill in the blank: For a power series centred at a = 2 with radius R = 4, the open interval of convergence (before checking endpoints) is ( ___ , ___ ). (Answer: (minus 2, 6).)
True or false: If a series converges conditionally at an endpoint, that endpoint is included in the interval of convergence. (True. Convergence is convergence, whether absolute or conditional.)
Fill in the blank: When you simplify the ratio test limit for a power series centred at a, you always end up with L = |x minus a| divided by ___. (Answer: R, the radius of convergence.)
True or false: The centre a is always the midpoint of the interval of convergence. (True.)
Q: Find the interval of convergence of the power series sum from n = 1 to infinity of x^n / n.
A: Apply the ratio test: L = lim |x^(n+1)/(n+1)| / |x^n/n| = |x| times lim n/(n+1) = |x|. Set |x| < 1, giving the open interval (minus 1, 1) with R = 1. At x = 1: sum of 1/n (harmonic series, diverges). At x = minus 1: sum of (minus 1)^n / n (alternating harmonic series, converges by AST). I.O.C. = [minus 1, 1).
Q: A power series centred at a = minus 3 has interval of convergence [minus 7, 1). What is the radius of convergence?
A: The interval stretches from minus 7 to 1. The width is 8, so R = 4. Check: the centre minus 3 is exactly 4 units from each endpoint.
Q: After applying the ratio test to a power series, you find L = |x minus 2| / 5. What are the centre, the radius, and the open interval of convergence?
A: Centre a = 2, radius R = 5. Open interval: (2 minus 5, 2 plus 5) = (minus 3, 7). Endpoints x = minus 3 and x = 7 still need to be checked separately.
Q: You check the left endpoint of a power series and get the series sum from n = 1 to infinity of (minus 1)^n / n^2. Does this endpoint belong in the I.O.C.?
A: Yes. The series sum of (minus 1)^n / n^2 converges absolutely because sum of 1/n^2 is a convergent p-series (p = 2 > 1). Include the endpoint.
This material connects to Taylor and Maclaurin series, where you will construct power series representations for specific functions and need to state their intervals of convergence.
The convergence tests from earlier in the course (ratio test, p-series, alternating series test, comparison test) are not just prerequisites here; they are active tools you use at the endpoint-checking stage.
In later courses (differential equations, complex analysis), the radius of convergence takes on deeper significance: it corresponds to the distance to the nearest singularity of the function in the complex plane.
Interval of convergence, I.O.C., radius of convergence, ratio test for power series, endpoint convergence, absolute convergence, conditional convergence, alternating series test, p-series test, harmonic series, power series convergence, finding I.O.C., Calculus II, Calculus 2, Calc 2, XX123, series and sequences, convergence tests, open interval, bracket notation, centre of convergence