What Happens to Convergence When You Differentiate a Power Series?

Differentiation preserves the radius of convergence of a power series, but not the endpoints. A power series expansion calculator is useful at that point for checking the coefficients you start with, while the endpoint tests still have to be done separately. If your original series converges for |x - c| < R, the differentiated series also converges for |x - c| < R. That is the preserved part. The endpoints at x = c ± R, however, require separate testing because the derivative series can diverge at one or both endpoints even when the original series converged there. This is a specific, testable consequence of the theorem on term-by-term differentiation, and it is the core answer to the question: what happens to convergence when you differentiate a power series?

The ratio test or root test applied to the differentiated series yields the same limit as for the original series, guaranteeing the same radius R. For a series ∑ a_n (x - c)^n, the derivative series is ∑ n a_n (x - c)^(n-1). The factor n does not affect the limit in the ratio test because lim |(n+1) / n| = 1. The same holds for the root test, where the nth root of n approaches 1. So the radius of convergence is identical before and after differentiation. This is not an approximation or a rule of thumb; it is a theorem from the standard analysis of power series, stated in OpenStax Calculus Vol. 2, Section 5.6.

Take the original series ∑ x^n / n from n = 1 to ∞. Its center c = 0. The ratio test on a_n = 1/n gives lim |a_{n+1}/a_n| = lim n/(n+1) = 1, so R = 1. The interval of convergence, before endpoint testing, is (-1, 1). Now test the endpoints for this original series. At x = 1, the series becomes ∑ 1/n, the harmonic series, which diverges. At x = -1, the series becomes ∑ (-1)^n / n, an alternating harmonic series that converges conditionally. So the original series converges at x = -1 and diverges at x = 1.

Now differentiate term by term. The derivative series is ∑ n * (1/n) x^(n-1) = ∑ x^(n-1) from n = 1 to ∞, which equals ∑ x^k from k = 0 to ∞. This is a geometric series with first term 1 and common ratio x. Its radius of convergence is also R = 1, confirming the preservation. Now test its endpoints. At x = 1, the geometric series becomes ∑ 1, which diverges because the terms do not approach 0. At x = -1, it becomes ∑ (-1)^k, which also diverges because the terms alternate between 1 and -1 and never approach 0. So the derivative series converges only on the open interval (-1, 1), losing convergence at the left endpoint where the original series converged.

This is the critical point: the derivative series may lose convergence at endpoints even if the original series converged there. If you need to verify the calculation for a different series, you can use an expansion tool to check your radius and endpoint results, but you must understand that the tool's output for the derivative series' endpoints is not automatically the same as the original series' endpoints.

Endpoints require separate testing because the ratio and root tests are inconclusive at the boundary |x - c| = R. Both tests produce L = 1 at the endpoints, which does not tell you whether the series converges or diverges. This inconclusiveness is inherited by the derivative series: the ratio test on the derivative series also gives L = 1 at the endpoints. So you must apply other convergence tests, p-series, alternating series, comparison, or integral tests, directly to the endpoint series after substitution.

Why the Derivative Series Can Behave Differently at Endpoints

The derivative series has terms multiplied by n. This factor n can strengthen or weaken the convergence at the boundary. In the example above, the original series at x = -1 was an alternating harmonic series, which converges because the terms 1/n decrease to zero. The derivative series at x = -1 became ∑ (-1)^k, whose terms do not decrease at all. The factor n cancelled the 1/n, turning a convergent alternating series into a divergent geometric series. This is not a rare edge case; it is the standard behavior for many power series with polynomial coefficients.

The Failure Case

The most instructive failure case is the original series ∑ (-1)^n x^n / n. At x = 1, this series is ∑ (-1)^n / n, the alternating harmonic series, which converges conditionally. Its derivative series is ∑ (-1)^n x^(n-1). At x = 1, this becomes ∑ (-1)^n, which diverges by the nth-term test because the terms do not approach zero. So endpoint convergence of the original series does not guarantee endpoint convergence of the derivative series. If you rely on the original series' interval of convergence without rechecking, you will make an error.

When the coefficients involve factorials or exponents of n, the ratio test limit may not exist, or the algebra may be messy. In such cases, use the root test. For a power series ∑ a_n x^n, the root test computes L = lim sup |a_n|^(1/n). The radius R = 1 / L. For the derivative series ∑ n a_n x^(n-1), the root test becomes lim sup |n a_n|^(1/n). Since lim sup n^(1/n) = 1, the radius remains R. This is the Cauchy-Hadamard theorem in action: the factor n does not change the lim sup. So the root test is the more general method, and it confirms radius preservation without requiring the ratio test limit to exist.

Consider the original series ∑ x^n / n^2 from n = 1 to ∞. The ratio test on a_n = 1/n^2 gives R = 1. At x = 1, the series becomes ∑ 1/n^2, a p-series with p = 2, which converges. At x = -1, the series becomes ∑ (-1)^n / n^2, which converges absolutely because the absolute series is ∑ 1/n^2. So both endpoints converge for the original series. Differentiate: the derivative series is ∑ n * (1/n^2) x^(n-1) = ∑ (1/n) x^(n-1) from n = 1 to ∞, or ∑ x^k / (k+1) from k = 0 to ∞. Its radius is still 1. At x = 1, the derivative series becomes ∑ 1/(k+1), the harmonic series, which diverges. At x = -1, it becomes ∑ (-1)^k / (k+1), an alternating series that converges conditionally by the alternating series test. So one endpoint converged in the original series but diverged in the derivative series.

If you had assumed that the derivative series inherits both endpoints, you would incorrectly claim the derivative series converges at x = 1. This is why each endpoint must be tested from scratch after differentiation.

The theorem that differentiation preserves the radius of convergence is proven by comparing the ratio test limits of the original and derivative series. For the original series, the ratio test limit is L = lim |a_{n+1} / a_n|. For the derivative series, the limit is lim |(n+1) a_{n+1} / (n a_n)| = L * lim (n+1)/n = L. Since the limit of (n+1)/n is 1, the two limits are identical, so the radius is identical. The same logic applies to the root test. This is the reason the open interval (c - R, c + R) is preserved: inside this open interval, the series converges absolutely, and differentiation does not break absolute convergence.

What the Theorem Does Not Cover

The theorem does not cover the endpoints. At |x - c| = R, the ratio test returns L = 1, which is inconclusive. The proof relies on the ratio test producing a limit strictly less than 1 inside the open interval, but at the endpoints it produces exactly 1. So the theorem has no statement about endpoints. The endpoints are outside its jurisdiction, and you must test them with other convergence tests.