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#.