Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 15

A power series centered at c=0 is follows the formula:
sum_(n=0)^oo a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+...
The given function f(x)= 2/(1-x^2) centered at c=0 resembles the power series:
(1+x)^k = sum_(n=0)^oo (k(k-1)(k-2)...(k-n+1))/(n!) x ^n
or
(1+x)^k = 1+kx +(k(k-1))/(2!)x^2+(k(k-1)(k-2))/(3!)x^3+(k(k-1)(k-2)(k-3))/(4!)x^4+...
Apply Law of exponents: 1/x^n = x^(-n) , we get: f(x)= 2(1-x^2) ^(-1)
Apply the aforementioned formula for power series on (1-x^2) ^(-1) or (1+(-x^2)) ^(-1) , we may replace "x " with "-x^2 " and "k " with "-1 ". We let:
(1+(-x^2))^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) (-x^2) ^n
=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)(-1)^nx ^(2n)
=1+(-1)(-1)^1x ^(2*1) +(-1(-2))/(2!)(-1)^2x ^(2*2)+(-1(-2)(-3))/(3!)(-1)^3x ^(2*3)+(-1(-2)(-3)(-4)/(4!)(-1)^4x ^(2*4)+...
=1+(-1)*(-1)x^2 +2/2*1*x ^4 +(-6)/6*(-1)*x^6+24/24*1*x^8+...
=1+1x^2 +2/2x ^4 +6/6x^6+24/24x^8+...
=1+x^2 +x ^4 +x^6+x^8+...
Applying (1+(-x^2))^(-1) =1+x^2 +x ^4 +x^6+x^8+... , we get:
2*(1+(-x^2))^(-1) =2*[ 1+x^2 +x ^4 +x^6+x^8+...]
=2+2x^2 +2x ^4 +2x^6+2x^8+...
= sum_(n=0)^oo 2x^(2n)
The power series of the function f(x)=2/(1-x^2) centered at c=0 is:
2/(1-x^2) =sum_(n=0)^oo 2x^(2n)
or
2/(1-x^2) =2+2x^2 +2x ^4 +2x^6+2x^8+...
To determine the interval of convergence, we may apply geometric series test wherein the series sum_(n=0)^oo a*r^n is convergent if |r|lt1 or -1 ltrlt 1 . If |r|gt=1 then the geometric series diverges.
The series sum_(n=0)^oo 2x^(2n) is the same as sum_(n=0)^oo 2(x^2)^n .
By comparing sum_(n=0)^oo2(x^2)^n with sum_(n=0)^oo a*r^n , we determine: r =x^2 .
Apply the condition for convergence of geometric series: |r|lt1 .
|x^2|lt1
|x|^2lt1
For xgt=0 , we replace |x| with x .
x^2lt1
Applying (f(x))^2lta then f(x)ltsqrt(a) and f(x) gt-sqrt(a) .
Let f(x) =x and a =1 then +- sqrt(a)=+- sqrt(1) = +-1 .
The interval will be xlt1 and xgt-1 and combine as -1ltxlt1 .
For xlt0 , we replace |x| with -x.
(-x)^2lt1
x^2 lt1
The interval will also be -1ltxlt1 .
Combining the ranges from xlt0 and or xgt=0 , we get the interval for convergence as: -1ltxlt1 .
Check the convergence at endpoints that may satisfy |x^2|=1 .
Let x=-1 on sum_(n=0)^oo 2(x^2)^n , we get:
sum_(n=0)^oo2((-1)^2)^n=sum_(n=0)^oo 2(1)^n
Using geometric series test, the r =1 satisfy |r| gt=1 . Thus, the series diverges at x=-1 .
Let x=1 on sum_(n=0)^oo 2(x^2)^n , we get:
sum_(n=0)^oo2(1^2)^n=sum_(n=0)^oo 2(1)^n
Using geometric series test, the r =1 satisfy |r| gt=1 . Thus, the series diverges at x=1 .
Thus, the power series sum_(n=0)^oo 2(x^2)^n or sum_(n=0)^oo2x^(2n) has an interval of convergence: -1 ltxlt1 .