# How do you find a power series representation for f(x)= 1/(1+x) and what is the radius of convergence?

Oct 5, 2015

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n}$ with radius of convergence $1$

#### Explanation:

Start writing out a power series which when multiplied by $\left(1 + x\right)$ gives $1$...

$1 = \left(1 + x\right) \left(1 - x + {x}^{2} - {x}^{3} + {x}^{4} - \ldots\right)$

We choose each successive term to cancel out the extraneous term left over by the previous ones.

Then writing it out formally...

$\left(1 + x\right) {\sum}_{n = 0}^{N} {\left(- 1\right)}^{n} {x}^{n}$

$= {\sum}_{n = 0}^{N} {\left(- 1\right)}^{n} {x}^{n} + x {\sum}_{n = 0}^{N} {\left(- 1\right)}^{n} {x}^{n}$

$= {\sum}_{n = 0}^{N} {\left(- 1\right)}^{n} {x}^{n} - {\sum}_{n = 1}^{N + 1} {\left(- 1\right)}^{n} {x}^{n}$

$= {\left(- 1\right)}^{0} {x}^{0} - {\left(- 1\right)}^{N + 1} {x}^{N + 1} = 1 - {\left(- x\right)}^{N + 1}$

So if $\left\mid x \right\mid < 1$, then ${\left(- x\right)}^{N + 1} \to 0$ as $N \to \infty$ and we find

$\left(1 + x\right) {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{n} = 1$

Conversely, if $\left\mid x \right\mid \ge 1$ then ${\lim}_{N \to \infty} {\left(- x\right)}^{N + 1}$ does not exist and the sum does not converge.

So the radius of convergence is $1$