# How do you find the power series for f(x)=1/(1+x^2)^2 and determine its radius of convergence?

Jan 9, 2018

The power series is:

f(x) =sum_(n=0)^oo (-1)^(n-1) nx^(2(n-1)

This gives a radius of convergence of: $| x | < 1$

#### Explanation:

Start with the form of the power series for a geometric series:

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

Now find the derivative of this:

$\frac{d}{\mathrm{dr}} {\sum}_{n = 0}^{\infty} {r}^{n} = {\sum}_{n = 0}^{\infty} n {r}^{n - 1}$

Also:

$\frac{d}{\mathrm{dr}} \frac{1}{1 - r} = \frac{1}{1 - r} ^ 2$

Now replace $r \to - {x}^{2}$ so that:

$\frac{1}{1 - r} ^ 2 \to \frac{1}{1 + {x}^{2}} ^ 2$

and also therefore that:

${\sum}_{n = 0}^{\infty} n {r}^{n - 1} \to {\sum}_{n = 0}^{\infty} n {\left(- {x}^{2}\right)}^{n - 1} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n - 1} n {x}^{2 \left(n - 1\right)}$

So we have found that the power series is therefore:

1/(1+x^2)^2 =sum_(n=0)^oo (-1)^(n-1) nx^(2(n-1)

From this the first few terms are:

$f \left(x\right) = 1 - 2 {x}^{2} + 3 {x}^{4} - 4 {x}^{6} + \ldots$

which if you check is in agreement with the Taylor expansion around $x = 0$.

So we can now find the radius of convergence (i.e. the values of $x$ for which this sum will converge). The radius of convergence can be found by:

If ${A}_{n}$ is the $n$th term of the series then by application of the ratio test of convergence: the series will converge if:

${\lim}_{n \to \infty} | {A}_{n + 1} \frac{|}{|} {A}_{n} | < 1$

In our case: ${A}_{n} = {\left(- 1\right)}^{n} \left(n + 1\right) {x}^{2 n}$ so for convergence:

${\lim}_{n \to \infty} | {\left(- 1\right)}^{\left(n - 1\right) + 1} \left(n + 1\right) {x}^{2 \left(\left(n - 1\right) + 1\right)} \frac{|}{|} {\left(- 1\right)}^{n - 1} n {x}^{2 \left(n - 1\right)} | < 1$

$\to {\lim}_{n \to \infty} | {\left(- 1\right)}^{n} \left(n + 1\right) {x}^{2 n} \frac{|}{|} {\left(- 1\right)}^{n - 1} n {x}^{2 \left(n - 1\right)} | < 1$

$\to {\lim}_{n \to \infty} | - 1 \frac{n + 1}{n} {x}^{2} | = {\lim}_{n \to \infty} \frac{n + 1}{n} | {x}^{2} | < 1$

Evaluating:

${\lim}_{n \to \infty} \frac{n + 1}{n} = {\lim}_{n \to \infty} \frac{n}{n} = 1$

$\to 1 | {x}^{2} | < 1 \to | x {|}^{2} < 1 \therefore | x | < 1$

So finally we arrive at the radius of convergence of $1$.