Differentiating and Integrating Power Series

Work inside the radius

A power series can be differentiated and integrated term by term inside its open convergence interval. The resulting series has the same radius, although endpoint behavior can change. On any closed subinterval strictly inside the radius, these operations have the convergence control needed to justify them.

Starting from $1/(1-x)=\sum_{n=0}^\infty x^n$ for $|x|<1$ , differentiation yields

\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}.

Multiplication by $x$ then gives $x/(1-x)^2=\sum_{n=1}^\infty nx^n$ . Index shifts change the displayed starting index and exponent; keep both aligned.

Visual guide

VISUAL GUIDEDifferentiating a series changes its coefficients

1/(1 − x)²
1 + 2x + ⋯ + 6x⁵

Differentiating the geometric series gives 1/(1 − x)² = 1 + 2x + 3x² + ⋯ for |x| < 1. The plotted truncation follows the target near zero and loses accuracy near the singularity. Endpoint behavior must be checked again.

Worked example: integrate from a known base point

Integrate the geometric series from $0$ to $x$ :

-\ln(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=\sum_{n=1}^\infty\frac{x^n}{n},\quad |x|<1.

The lower endpoint fixes the integration constant because both sides vanish at $x=0$ . An indefinite termwise integral needs an arbitrary constant just like any other indefinite integral.

At $x=-1$ , the new series converges conditionally, while the original geometric series there diverges. Endpoint inclusion must be checked again after integrating. The interior identity alone is not a license to substitute arbitrary boundary values without an additional endpoint argument.

Worked example: substitution before integration

Replace $x$ by $-t^2$ in the geometric identity to obtain $1/(1+t^2)=\sum_{n=0}^\infty(-1)^nt^{2n}$ for $|t|<1$ . Integrating from zero gives

\arctan x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{2n+1},\quad |x|<1.

At $x=1/2$ , the first two terms give $1/2-(1/2)^3/3=11/24$ . The alternating remainder is at most $(1/2)^5/5=1/160$ because the term magnitudes decrease.

Practice

Expand $1/(1-2x)$ and state its radius.
Obtain a series for $x^2/(1-x)$ .
Differentiate $\sum_{n=0}^\infty x^{2n}$ inside its radius.

Show worked solutions

$\sum_{n=0}^\infty2^nx^n$ , with radius $1/2$ .
$\sum_{n=0}^\infty x^{n+2}$ for $|x|<1$ .
$\sum_{n=1}^\infty2nx^{2n-1}=2x/(1-x^2)^2$ , for $|x|<1$ .