f(x) = 1/(3-x) , c=1 Find a power series for the function, centered at c and determine the interval of convergence.

To determine the power series centered at c, we may apply the formula for Taylor series:
f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n
or
f(x) =f(c)+f'(c)(x-c) +(f''(c))/(2!)(x-c)^2 +(f^3(c))/(3!)(x-c)^3 +(f'^4(c))/(4!)(x-c)^4 +...
To list the f^n(x) for the given function f(x)=1/(3-x) centered at c=1 , we may apply Law of Exponent: 1/x^n = x^-n  and  Power rule for derivative: d/(dx) x^n= n *x^(n-1) .
f(x) =1/(3-x)
      = (3-x)^(-1)
Let u =3-x then (du)/(dx) = -1
d/(dx) c*(3-x)^n = c *d/(dx) (3-x)^n
                          = c *(n* (3-x)^(n-1)*(-1)
                          = -cn(3-x)^(n-1)
f'(x) =d/(dx) (3-x)^(-1)
           =-(-1)(3-x)^(-1-1)
            =(3-x)^(-2) or 1/(3-x)^2
f^2(x) =d/(dx) (3-x)^(-2)
            =-(-2)(3-x)^(-2-1)
            =2(3-x)^(-3) or 2/(3-x)^3
f^3(x) =d/(dx)2(3-x)^(-3)
            =-2(-3)(3-x)^(-3-1)
            =6(3-x)^(-4) or 6/(3-x)^4
f^4(x) =d/(dx)6(3-x)^(-4)
           =-6(-4)(3-x)^(-4-1)
           =24(3-x)^(-5) or 24/(3-x)^5
Plug-in x=1 for each f^n(x) , we get:
f(1)=1/(3-1) =1/2
f'(1)=1/(3-1)^2 = 1/4
f^2(1)=2/(3-1)^3 =1/4
f^3(1)=6/(3-1)^4 = 3/8
f^4(1)=24/(3-1)^5 = 3/4
Plug-in the values on the formula for Taylor series, we get:
1/(3-x) = sum_(n=0)^oo (f^n(1))/(n!) (x-1)^n
=f(1)+f'(1)(x-1) +(f^2(1))/(2!)(x-1)^2 +(f^3(1))/(3!)(x-1)^3 +(f^4(1))/(4!)(x-1)^4 +...
 
=1/2+1/4(x-1) +(1/4)/(2!)(x-1)^2 +(3/8)/(3!)(x-1)^3 +(3/4)/(4!)(x-1)^4 +...
=1/2+1/4(x-1) +(1/4)/2(x-1)^2 +(3/8)/6(x-1)^3 +(3/4)/24(x-1)^4 +...
=1/2+1/4(x-1) + 1/8(x-1)^2 +1/16(x-1)^3 +1/32(x-1)^4 +...
=sum_(n=1)^oo ((x-1)/2)^n
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.
By comparing sum_(n=0)^oo ((x-1)/2)^n or  sum_(n=0)^oo 1*((x-1)/2)^n with  sum_(n=0)^oo a*r^n , we determine: r = (x-1)/2 .
Apply the condition for convergence of geometric series: |r|lt1 .
|(x-1)/2|lt1
-1lt(x-1)/2lt1
Multiply each sides by 2:
-1*2lt(x-1)/2*2lt1*2
-2ltx-1lt2
Add 1 on each sides:
-2+1ltx-1+1lt2+1
-1ltxlt3
Thus, the power series  of the function f(x) = 1/(3-x) centered at c=1 is sum_(n=1)^oo ((x-1)/2)^n  with an interval of convergence: -1ltxlt3 .