Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 21

Binomial series is an example of an infinite series. When it is convergent at |x|lt1 , we may follow the sum of the binomial series as (1+x)^k where k is any number. The formula will be:
(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+...
To evaluate the given function f(x) = 1/sqrt(4+x^2) , we may apply 4+x^2=4(1+x^2/4) .
The function becomes:
f(x) =1/sqrt(4(1+x^2/4))
f(x) =1/(2sqrt(1+x^2/4))
Apply radical property: sqrt(x) = x^(1/2) . The function becomes:
f(x) =1/(2(1+x^2/4)^(1/2))
Apply Law of Exponents: 1/x^n = x^(-n) to rewrite the function as:
f(x) =1/2(1+x^2/4)^(-1/2)
or f(x)=1/2(1+x^2/4)^(-0.5)
Apply the aforementioned formula on (1+x^2/4)^(-0.5) by letting:
x=x^2/4 and k =-0.5.
(1+x^2/4)^(-0.5) = sum_(n=0)^oo (-0.5(-0.5-1)(-0.5-2) ...(-0.5-n+1))/(n!) (x^2/4)^n
= sum_(n=0)^oo (-0.5(-1.5)(-2.5) ...(-0.5-n+1))/(n!) x^(2n)/4^n
= 1 + (-0.5)*x^(2*1)/4^1+ (-0.5(-1.5))/(2!) *x^(2*2)/4^2+ (-0.5(-1.5)(-2.5))/(3!)*x^(2*3)/4^3 +(-0.5(-1.5)(-2.5)(-3.5))/(4!)*x^(2*4)/4^4+...
= 1 -0.5x^2/4+ 0.75/2* x^4/16-1.875/6x^6/64 +6.5625/24x^8/256+...
= 1 -x^2/8+ (3x^4)/128-(5x^6)/1024 +(35x^8)/32768+...
Applying (1+x^2/4)^(-0.5) =1 -x^2/8+ (3x^4)/128-(5x^6)/1024 +(35x^8)/32768+... , we get:
1/2(1+x^2/4)^(-0.5) = 1/2[1 -x^2/8+ (3x^4)/128-(5x^6)/1024 +(35x^8)/32768+...]
=1/2 -x^2/16+ (3x^4)/256-(5x^6)/2048+(35x^8)/65536+...
Therefore, the Maclaurin series for the function f(x) =1/sqrt(4+x^2) can be expressed as:
1/sqrt(4+x^2)=1/2 -x^2/16+ (3x^4)/256-(5x^6)/2048+(35x^8)/65536+...