The geometric series formula, read backward, says that a function equals a power series: $$\frac{1}{1 - x} = 1 + x + x^{2} + x^{3} + \cdots,$$ $$(|x| < 1).$$ Many other functions can be rewritten in the form $\frac{c}{1 - u}$, and then their series come for free. This builds directly on power series and intervals of convergence.
Por qué funciona
For $|x| < 1$, the partial sums of $\sum x^{n}$ are $\frac{1 - x^{N}}{1 - x}$, and $x^{N} \to 0$, so the series converges to $\frac{1}{1 - x}$. The identity holds only inside $(-1, 1)$: at $x = 2$ the left side is $-1$ while the right side diverges. Substituting any expression $u$ for $x$ keeps the identity true wherever $|u| < 1$, which also tells you the new interval of convergence.
Idea clave
$$\frac{1}{1 - u} = \sum_{n=0}^{\infty} u^{n} \quad\text{for } |u| < 1.$$
Rewrite the function as $\frac{c}{1 - u}$ (factor constants out of the denominator first), substitute, and read the interval from $|u| < 1$.
Ejemplo resuelto: Substitution
Find a power series for $\frac{1}{1 + x^{2}}$ and its interval of convergence.
- $\frac{1}{1 + x^{2}} = \frac{1}{1 - (-x^{2})}$, so $u = -x^{2}$.
- $$\begin{aligned} &\sum_{n=0}^{\infty}(-x^{2})^{n} \\ &= 1 - x^{2} + x^{4} - x^{6} + \cdots \\ &= \sum_{n=0}^{\infty}(-1)^{n}x^{2n}. \end{aligned}$$
- Valid when $|-x^{2}| < 1$, that is $-1 < x < 1$.
Respuesta: $\sum(-1)^{n}x^{2n}$ on $(-1, 1)$
Ejemplo resuelto: Factor out a constant
Find a power series for $\frac{3}{2 - x}$ centered at 0.
- Factor 2 from the denominator: $\frac{3}{2 - x} = \frac{3}{2}\cdot\frac{1}{1 - \frac{x}{2}}$.
- $= \frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{x}{2}\right)^{n} = \sum_{n=0}^{\infty}\frac{3}{2^{n+1}}x^{n}$.
- Valid when $\left|\frac{x}{2}\right| < 1$: $|x| < 2$, so $R = 2$.
Respuesta: $\sum\frac{3x^{n}}{2^{n+1}}$, $R = 2$
Error común
Taking the radius to be 1 every time. The radius comes from $|u| < 1$: for $\frac{3}{2 - x}$, $u = \frac{x}{2}$, so $R = 2$. A quick check: the radius is the distance from the center to the nearest point where the function blows up (here $x = 2$).
Ejemplo resuelto: Multiply by a power of x
Find a power series for $\frac{x}{1 - x^{2}}$.
- $\frac{1}{1 - x^{2}} = 1 + x^{2} + x^{4} + \cdots$ for $|x| < 1$.
- Multiply by $x$: $x + x^{3} + x^{5} + \cdots = \sum_{n=0}^{\infty}x^{2n+1}$, still for $|x| < 1$.
Respuesta: $\sum x^{2n+1}$
The same trick works around other centers. For $\frac{1}{x}$ near $x = 1$, write $$\begin{aligned} \frac{1}{x} &= \frac{1}{1 - (1 - x)} \\ &= \sum_{n=0}^{\infty}(1 - x)^{n} \\ &= \sum_{n=0}^{\infty}(-1)^{n}(x - 1)^{n}, \end{aligned}$$ valid for $|x - 1| < 1$, that is $0 < x < 2$.