Radius and Interval of Convergence

Before you can find where a power series converges, you need to separate two ideas. The radius of convergence (R) is a single nonnegative number, possibly 0 or infinity, that tells you the size of the interval of x-values where the series converges. The interval of convergence is the actual set of x-values, and it is always centered at the series' center a. A common mistake is to assume the interval is always open because the ratio test gives you a strict inequality, but endpoint convergence is a separate question that requires its own test. The series might converge absolutely, converge conditionally, or diverge at each endpoint, and each endpoint is independent of the other.

OpenStax Calculus Volume 2, Chapter 6 (Sections 6.1 and 6.2) defines the radius and interval and emphasizes that the convergence is one of three cases: R = 0 (converges only at x = a), R = ∞ (converges for all real x), or a finite positive number. The radius is a real number; higher is better because it means the series works over a wider interval of x, but it does not guarantee accuracy near the endpoints. When you use a power series to approximate a function, you must stay strictly inside the interval unless you have proven endpoint convergence, and even then, conditional convergence at an endpoint is fragile. The alternating series test may give you convergence, but the series of absolute values diverges, so rearranging terms can change the sum.

Finding the Radius with the Ratio Test

The power series method is what you use to find R in most homework problems. The procedure is mechanical: given a series ∑ c_n (x − a)^n, compute the limit L = lim_{n→∞} |a_{n+1} / a_n|, where a_n = c_n (x − a)^n. The series converges if L < 1, which translates to |x − a| < R, where R = lim_{n→∞} |c_n / c_{n+1}| if that limit exists. If L = 1, the test is inconclusive, and you must use a different test. The most common student mistake is to forget that the ratio test gives you R, not the interval, and then to report the open interval without testing endpoints.

Here is a worked example where the ratio test fails at the endpoint. The interval of convergence is [−1, 1), not (−1, 1). If you stop at R = 1, you lose the endpoint x = −1, which is a real loss because the series converges there. In another case, the interval is the single point {0}. OpenStax Chapter 6 calls this the R = 0 case, and you should not test endpoints because there are none. The ratio test is your first tool, but when the limit equals 1, you must switch to a different test, such as the alternating series test for endpoints or a comparison test. Do not skip the endpoint step, because the ratio test cannot tell you the answer there.

When the series has terms with powers that depend on n, the root test often works faster. This is often written as 1/R = lim sup |c_n|^{1/n}. The root test made finding R trivial because the exponent was linear in n. Now consider a case where the root test is the only practical route. The limit as n → ∞ is ∞ for any x ≠ 0, so L = ∞, and the series diverges for all x ≠ 0. At x = 0, every term after the first is 0, so the series converges. The radius is R = 0, and the interval is the single point {0}.

Another example: ∑ (x^n) / n^n. The root test gives you the answer in one line, whereas the ratio test would require you to compute lim |x^{n+1}/(n+1)^{n+1}| · |n^n / x^n| = |x| lim (n/(n+1))^n · (1/(n+1)) = 0, which is doable but messier. But remember, the root test also fails when L = 1; then you must test endpoints separately, exactly as with the ratio test.

After you have R from the ratio or root test, the question becomes how to find interval of convergence, and the answer always involves testing the endpoints x = a − R and x = a + R. The interval of convergence is the set of all x for which the series converges, and it is always centered at a, so the endpoints are symmetric around a. You must test each endpoint individually. Each endpoint is independent: you might get convergence at both, one, or neither. The radius is the same (R = 1) in both cases, but the interval differs because the endpoint behavior differs.

Here is a worked example where both endpoints converge. At x = 1, ∑ n diverges (terms do not tend to 0). The endpoint tests are where most homework errors happen: students find R, then write the open interval, forgetting that the series might converge at an endpoint. The alternating series test is useful when the endpoint series is alternating, and it does not require absolute convergence. If the endpoint series is not alternating, use a p-series or comparison. Never use the ratio test at an endpoint, because it will always give L = 1 there, which is inconclusive. Instead, switch to a different test, and always state the final interval with the correct brackets.

Three Outcomes for Endpoint Behavior

Endpoint convergence is not a single thing. Outcome one: both endpoints converge. Outcome two: one endpoint converges, the other diverges. For example, at x = −1, ∑ (−1)^n n also diverges because the terms do not tend to 0. These three outcomes are exhaustive for a power series; you will never see one endpoint converge absolutely and the other conditionally, because absolute convergence at one endpoint implies absolute convergence at the other only if the series is symmetric, which it is not necessarily. If either fails, the test is inconclusive, and you must use another test, such as the divergence test (if terms do not tend to 0) or a comparison test.

Here is a fifth worked example that combines all three outcomes in one problem. This is why you must test each endpoint individually, not assume a pattern. The only way to know is to plug in and apply a test, and the alternating series test is the one you will use most often for conditional convergence.

You might be tempted to use an interval of convergence calculator to check your homework, but you need to know its limits. Most online calculators, including the one discussed here, take a power series and a specific x-value, then evaluate the series numerically at that point. They do not compute the radius or the interval for you. The calculator’s page claims to compute the “radius and interval of convergence,” but in reality, it only evaluates the series at one x at a time, so you must already know R beforehand. The page’s how-to guide shows you how to enter a known series and a point, and it displays the partial sum and an error estimate, but it gives no formula for the remainder or how to bound it. The FAQ duplicates other pages’ content without addressing the actual confusion, such as why endpoints matter.

The claimed versus real is stark: the title promises a computation it does not perform. The only legitimate source for the math is OpenStax Calculus Volume 2, Chapter 6, not the calculator’s output. Here is what the calculator can do for you: it can check a single point. If you have already found R and tested the endpoints by hand, you can verify that the series converges at x = a + R by plugging that value in and seeing if the partial sums approach a limit. But the calculator will not tell you whether the alternating series test applies. For homework or an exam, you must show the ratio test to find R, then test endpoints with a hand test. The calculator’s error columns are misleading because they show the difference between the partial sum and the calculator’s own approximation, not the true error bound. The page’s claim that it “computes the radius and interval” is false, and the page’s own FAQ admits it only evaluates at a point. So do not trust the title. If you need a quick check, evaluate the series at a point you know is inside, like the center, and see if the partial sums settle. The one thing the calculator is good for is checking your final answer: if you think the interval is [−1, 1) and you plug in x = 0.9, the partial sums should converge; if they diverge, you made an error.

The cause of most errors is memorizing the formula without understanding that the ratio test gives a strict inequality, but the series might still converge at the boundary. Another mistake is misapplying the ratio test by not simplifying the ratio fully, leading to a wrong R. For example, if you have ∑ (x^n) / (n^2), and you write |a_{n+1}/a_n| = |x| · n^2 / (n+1)^2, you must take the limit, which is |x|, so R = 1. Term-by-term differentiation and integration are valid inside the interval, and the radius stays the same, but the endpoints may change. OpenStax Chapter 6 states that you can differentiate or integrate term-by-term within the interval of convergence, and the radius is preserved, but you must re-test the endpoints because the new series may converge or diverge there. If you integrate without re-testing, you will report the wrong interval.

Another mistake is treating the partial sum as exact. When you approximate a function with a power series, you are truncating the series, and the error is the remainder. The Taylor remainder formula, given in OpenStax Chapter 6, is R_N(x) = f^{(N+1)}(z) (x−a)^{N+1} / (N+1)! for some z between a and x, and the Lagrange error bound uses M = max |f^{(N+1)}(z)|. If you do not state the error, you are not doing the problem correctly. For homework, always write “for x in the interval” and then give the interval you found. A fifth mistake is using the ratio test when the terms do not involve factorials or powers, such as ∑ 1/n^2, which is a p-series, not a power series, the ratio test gives L = 1, inconclusive, so you must use the p-series test. Because the limit is 0, but you already know it converges, so do not waste time. The honest caveat: power series are not magic. So when you finish a problem, ask yourself: did I test both endpoints? If you answered no to any, go back and fix it.