Differentiate a Power Series Term by Term Explained

Q: Why does term-by-term differentiation preserve the radius of convergence?

The ratio test applied to the differentiated series produces the same limit as the original. For a series Σ cₙ(x-a)ⁿ, the ratio test limit for the differentiated series Σ n·cₙ(x-a)ⁿ⁻¹ is lim |(n+1)cₙ₊₁ / n·cₙ|, which equals lim |cₙ₊₁ / cₙ| because the factor n+1 over n approaches 1. That limit is exactly the reciprocal of R for the original series.

Q: Can the differentiated series converge at an endpoint where the original series diverges?

It is possible. For example, the original series may diverge at an endpoint, but the differentiated series converges there. Endpoint behavior is independent, no theorem states that convergence of the original at an endpoint implies convergence of the derivative, or vice versa. Each series must be tested on its own terms at the endpoints.

Q: What test do I use for the endpoint check on the differentiated series?

The ratio test returns 1 at the endpoints, so it is inconclusive. Use the p-series test, the alternating series test, or direct comparison. For a differentiated series that alternates at an endpoint, the alternating series estimation theorem can bound the error. For a p-series-like form, compare to 1/nᵖ.

Q: Is term-by-term differentiation valid if the original series converges only conditionally at an endpoint?

The theorem applies inside the open interval of convergence, not at the endpoints. Conditional convergence at an endpoint does not extend to the derivative. The example Σ xⁿ/n at x=-1 shows this: the original converges conditionally, but the differentiated series diverges.

Q: What should I do if the differentiated series diverges at an endpoint I need?

You cannot use the term-by-term series to represent the derivative at that point. Use another method: compute the derivative of the closed form if one exists, or evaluate the derivative as a limit using the original series and the definition of the derivative. The differentiated series is only valid for |x-a|<R, and you must respect that boundary.

Differentiate a Power Series Term by Term

You have a power series that represents a function inside its radius of convergence, and you need its derivative. The fastest route is to differentiate a power series term by term, treat each term as a simple polynomial and apply the power rule. The theorem guarantees this works for every x inside the open interval of convergence, |x-a|<R, and the differentiated series has the same radius R as the original. That is the good news. The catch is at the endpoints: x = a ± R, where the differentiated series can diverge even when the original series converges. You must check those endpoints separately, and that check is not optional.

A Worked Example

Pick a concrete series to see the mechanism. Start with the geometric series f(x) = 1/(1-x) = Σ xⁿ for |x|<1. Differentiate term by term: the derivative f'(x) = 1/(1-x)² becomes Σ n xⁿ⁻¹. Write out the first few terms: for n=1 you get 1·x⁰ = 1; for n=2 you get 2x; for n=3 you get 3x²; the sum is 1 + 2x + 3x² + ... . Test at a point inside the radius, say x=0.5. The closed form gives f'(0.5)=1/(0.5)²=4. The partial sum of the series: 1 + 2·0.5 + 3·0.25 = 1 + 1 + 0.75 = 2.75, which is not 4 yet, add more terms: the n=4 term is 4·0.125=0.5, making the sum 3.25; n=5 is 5·0.0625=0.3125, sum 3.5625; the series converges to 4 as the number of terms increases. The ratio test confirms the radius: for the original series, limit of |cₙ/cₙ₊₁| = 1, so R=1. For the differentiated series, the coefficients are n·1, so limit of |(n·1)/((n+1)·1)| = 1, same radius. The ratio test returns 1 at the endpoints, so you need another test. At x=1 the differentiated series is Σ n·1ⁿ⁻¹ = Σ n, which diverges by the nth-term test. At x=-1 the series is Σ n·(-1)ⁿ⁻¹, which also diverges because the terms do not approach zero. The original series diverges at x=1 and converges conditionally at x=-1, but the differentiated series diverges at both endpoints. That is the failure case: convergence at an endpoint of the original does not carry over.

Calculation Steps

When you need to differentiate a power series term by term for a function not in the standard list, the calculation follows three steps. First, confirm the radius of convergence R for the original series using a radius of convergence calculator or the ratio test. Second, write the differentiated series by multiplying each coefficient by its exponent and reducing the exponent by one: if the original term is cₙ(x-a)ⁿ, the derivative term is n·cₙ(x-a)ⁿ⁻¹. Third, test the endpoints x = a ± R on the differentiated series using the alternating series test or a p-series comparison. The differentiated series converges absolutely inside the open interval and uniformly on any closed subinterval [a-r, a+r] with r<R, but the endpoint behavior is independent and must be checked on the differentiated series itself, not assumed from the original.

For the specific series Σ xⁿ/n, the radius is 1. The original series converges conditionally at x=-1 by the alternating series test. Differentiate term by term to get Σ xⁿ⁻¹. At x=-1, the differentiated series becomes Σ (-1)ⁿ⁻¹, which diverges because the terms do not approach zero. The original converges at that endpoint; the derivative does not. That is the concrete failure case named in the research, and it shows why the endpoint check is not a formality.

Common Questions

Why does term-by-term differentiation preserve the radius of convergence?

The ratio test applied to the differentiated series produces the same limit as the original. For a series Σ cₙ(x-a)ⁿ, the ratio test limit for the differentiated series Σ n·cₙ(x-a)ⁿ⁻¹ is lim |(n+1)cₙ₊₁ / n·cₙ|, which equals lim |cₙ₊₁ / cₙ| because the factor n+1 over n approaches 1. That limit is exactly the reciprocal of R for the original series.

Can the differentiated series converge at an endpoint where the original series diverges?

It is possible. For example, the original series may diverge at an endpoint, but the differentiated series converges there. Endpoint behavior is independent, no theorem states that convergence of the original at an endpoint implies convergence of the derivative, or vice versa. Each series must be tested on its own terms at the endpoints.

What test do I use for the endpoint check on the differentiated series?

The ratio test returns 1 at the endpoints, so it is inconclusive. Use the p-series test, the alternating series test, or direct comparison. For a differentiated series that alternates at an endpoint, the alternating series estimation theorem can bound the error. For a p-series-like form, compare to 1/nᵖ.

Is term-by-term differentiation valid if the original series converges only conditionally at an endpoint?

The theorem applies inside the open interval of convergence, not at the endpoints. Conditional convergence at an endpoint does not extend to the derivative. The example Σ xⁿ/n at x=-1 shows this: the original converges conditionally, but the differentiated series diverges.

What should I do if the differentiated series diverges at an endpoint I need?

You cannot use the term-by-term series to represent the derivative at that point. Use another method: compute the derivative of the closed form if one exists, or evaluate the derivative as a limit using the original series and the definition of the derivative. The differentiated series is only valid for |x-a|<R, and you must respect that boundary.