# How to find the Maclaurin series and the radius of convergence for f(x)=1/(1+x)^2?

##### 1 Answer
Apr 29, 2018

$\frac{1}{1 + x} ^ 2 = {\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k + 1} k {x}^{k - 1}$ with $R = 1$

#### Explanation:

We start with the well known series:

$\frac{1}{1 - x} = {\sum}_{k = 0}^{\infty} {x}^{k} = 1 + x + {x}^{2} + {x}^{3} + \ldots$

Replace $x$ with $\left(- x\right)$:

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

Note that the derivative of $\frac{1}{1 + x} = {\left(1 + x\right)}^{-} 1$ is $- {\left(1 + x\right)}^{-} 2 = \frac{- 1}{1 + x} ^ 2$. We should then differentiate the series:

$\frac{- 1}{1 + x} ^ 2 = \frac{d}{\mathrm{dx}} {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{k} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{k} \frac{d}{\mathrm{dx}} {x}^{k} = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} \left(k {x}^{k - 1}\right)$

We can move the index of the sum to start at $k = 1$, since the term $k = 0$ results in an addend of just $0$.

We also multiply by a factor of $- 1$ to find the desired result:

$\frac{1}{1 + x} ^ 2 = - {\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k} k {x}^{k - 1} = {\sum}_{k = 1}^{\infty} {\left(- 1\right)}^{k + 1} k {x}^{k - 1}$

$= 1 - 2 x + 3 {x}^{2} - 4 {x}^{3} + \ldots$

Which has radius of convergence $R = 1$, since the original substitution of $- x$ for $x$ would not affect the radius, and neither would the differentiation or multiplication by a factor of $- 1$.