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

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

`sum_(n=0)^oor^n=1/(1-r)=1+r+r^2+r^3+...`

Now find the derivative of this:

`d/(dr)sum_(n=0)^oor^n = sum_(n=0)^oonr^(n-1)`

Also:

`d/(dr)1/(1-r)=1/(1-r)^2`

Now replace `r-> -x^2` so that:

`1/(1-r)^2-> 1/(1+x^2)^2`

and also therefore that:

`sum_(n=0)^oonr^(n-1)->sum_(n=0)^oo n(-x^2)^(n-1)=sum_(n=0)^oo (-1)^(n-1) nx^(2(n-1))`

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(x) = 1-2x^2+3x^4-4x^6+...`

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-> oo)|A_(n+1)|/|A_n| < 1`

In our case: `A_n = (-1)^n(n+1)x^(2n)` so for convergence:

`lim_(n->oo)|(-1)^((n-1)+1)(n+1)x^(2((n-1)+1))|/|(-1)^(n-1)nx^(2(n-1))|<1`

`->lim_(n->oo)|(-1)^n(n+1)x^(2n)|/|(-1)^(n-1)nx^(2(n-1))|<1`

`->lim_(n->oo)|-1(n+1)/nx^2|=lim_(n-> oo)(n+1)/n|x^2|<1`

Evaluating:

`lim_(n->oo)(n+1)/n = lim_(n->oo)n/n=1`

`-> 1 |x^2| < 1 -> |x|^2<1 therefore |x|<1`

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