Building new power series by substitution and integration

Understand the idea

A familiar power series can generate new ones through substitution, multiplication, differentiation, or integration. These operations let you represent functions without recomputing every derivative at a center. Track both the algebra and the region where the series representation is justified.

Begin with the geometric expansion and impose its strict convergence condition after any substitution. Multiplying by a polynomial changes powers and coefficients but does not generally remove the need to test boundary inputs. Make the first included term explicit.

Term-by-term differentiation and integration are valid inside the original radius of convergence and preserve that radius. Endpoint behavior can change, so retest endpoints after the operation. A constant of integration is determined by the function value at a convenient input.

\frac1{1-u}=\sum_{n=0}^{\infty}u^n\quad(|u|<1),\qquad \ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n}\quad(-1<x\le1)

A method you can apply

Choose a familiar series and rewrite the target function to match it.
Perform the substitution or termwise operation while keeping coefficients and exponents consistent.
Find the interior domain, test endpoints independently, and determine any added constant.

Worked example

Find a power series for $\displaystyle x^2/(1+3x)$ and its interval of convergence.

Rewrite the denominator as $\displaystyle 1-(-3x)$ and use the geometric series.
Multiplying by $\displaystyle x^2$ gives $\displaystyle \sum_{n=0}^{\infty}(-3)^n x^{n+2}=x^2-3x^3+9x^4-27x^5+\cdots$ .
The interior condition is $\displaystyle |3x|<1$ , so $\displaystyle -1/3<x<1/3$ .
At either endpoint, the term magnitudes are $\displaystyle 1/9$ and do not approach zero.

The series is

\displaystyle \sum_{n=0}^{\infty}(-3)^n x^{n+2}

, with interval

\displaystyle (-1/3,1/3)

.

Try it yourself

Integrate the geometric series for $\displaystyle 1/(1+x)$ from $\displaystyle 0$ to $\displaystyle x$ to obtain a series for $\displaystyle \ln(1+x)$ and its interval.

Show solution and reasoning

Inside $\displaystyle |x|<1$ , termwise integration gives $\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{n+1}/(n+1)$ , with zero constant because $\displaystyle \ln1=0$ . At $\displaystyle x=1$ the alternating harmonic series converges, while at $\displaystyle x=-1$ the resulting negative harmonic series diverges. The interval is $\displaystyle (-1,1]$ .