Calculus of a Single Variable, Chapter 9, 9.9, Section 9.9, Problem 11

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)= 3/(3x+4) 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+...
For better comparison, we let 3x+4 = 4 ((3x)/4 + 1) . The function becomes:
f(x)= 3/4 ((3x)/4 + 1)
Apply Law of exponents: 1/x^n = x^(-n) .
f(x)= 3/4((3x)/4 + 1)^(-1)

Apply the aforementioned formula for power series on ((3x)/4 + 1)^(-1) , we may replace "x" with "(3x)/4 " and "k " with "-1 ". We let:
(1+(3x)/4)^(-1) = sum_(n=0)^oo (-1(-1-1)(-1-2)...(-1-n+1))/(n!) ((3x)/4) ^n
=sum_(n=0)^oo (-1(-2)(-3)...(-1-n+1))/(n!)((3x)/4) ^n
=1+(-1)((3x)/4) +(-1(-2))/(2!)((3x)/4)^2+(-1(-2)(-3))/(3!)((3x)/4)^3+(-1(-2)(-3)(-4)/(4!)((3x)/4)^4+...
=1-(3x)/4 +(2)/2((3x)/4)^2- 6/6((3x)/4)^3+24/24((3x)/4)^4+...
=1-(3x)/4 +((3x)/4)^2- ((3x)/4)^3+((3x)/4)^4+...
=1-(3x)/4 +(9x^2)/16- (27x^3)/64+(81x^4)/256+...
Applying (1+(3x)/4)^(-1) =1-(3x)/4 +(9x^2)/16- (27x^3)/64+(81x^4)/256+... we get:
3/4((3x)/4 + 1)^(-1)= 3/4*[1-(3x)/4 +(9x^2)/16- (27x^3)/64+(81x^4)/256+...]
=3/4-(9x)/16 +(27x^2)/64- (81x^3)/256+(243x^4)/1024+...
= sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n
The power series of the function f(x)=3/(3x+4) centered at c=0 is:
3/(3x+4)=sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n
or
3/(3x+4)=3/4-(9x)/16 +(279x^2)/64- (81x^3)/256+(243x^4)/1024+...
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.
Applying (3/4)^(n+1) = (3/4)^n * (3/4) on the series sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n , we get:
sum_(n=0)^oo (-1)^n(3/4)^n(3/4)x^n =sum_(n=0)^oo(3/4) (-(3x)/4)^n
By comparing sum_(n=0)^oo(3/4) (-(3x)/4)^n with sum_(n=0)^oo a*r^n , we determine:r =-(3x)/4 .
Apply the condition for convergence of geometric series: |r|lt1 .
|-(3x)/4|lt1
|-1| *|(3x)/4|lt1
1 *|(3x)/4|lt1
|(3x)/4|lt1
-1lt(3x)/4lt1
Multiply each sides by 4/3 :
-1*4/3lt(3x)/4*4/3lt1*4/3
-4/3 ltxlt4/3
Check the convergence at endpoints that may satisfy |(3x)/4|=1 .
Let x=-4/3 on sum_(n=0)^oo(3/4) (-(3x)/4)^n , we get:
sum_(n=0)^oo(3/4) (-3/4*-4/3)^n=sum_(n=0)^oo(1)^n
Using geometric series test, the r =1 satisfy |r| gt=1 . Thus, the series diverges at x=-4/3 .
Let x=4/3 on sum_(n=0)^oo(3/4) (-(3x)/4)^n , we get:
sum_(n=0)^oo(3/4) (-3/4*4/3)^n=sum_(n=0)^oo(-1)^n
Using geometric series test, the r =-1 satisfy |r| gt=1 . Thus, the series diverges at x=-4/3 .
Thus, the power series sum_(n=0)^oo (-1)^n(3/4)^(n+1)x^n has an interval of convergence: -4/3 ltxlt4/3 .