Using the power rule we can determine the derivatives of #f(x)# of all orders:

#f(x) = (1-x)^(-1)#

#(df)/dx = (-1)(-1)(1-x)^(-2)= (1-x)^(-2)#

#(d^2f)/(dx^2) = (-1)(-2)(1-x)^(-3)= 2(1-x)^(-3)#

and we can easily see that:

#(d^nf)/(dx^n) =(n!) /(1-x)^(n+1)#

For #x=3# then the Taylor coefficients are:

#f^((n))(3)/(n!) = (n!)/( (1-3)^(n+1)(n!)) = (-1)^(n+1)/2^(n+1)#

And the Taylor series is:

#1/(1-x) = sum_(n=0)^oo (-1)^(n+1)(x-3)^n/2^(n+1)#

or:

#1/(1-x) = -1/2 sum_(n=0)^oo ((3-x)/2)^n#

We can obtain the same result directly by substituting #t=x-3# in the original expression:

#1/(1-x) = 1/(1-t-3) = -1/(t+2) = -1/2 1/(1+t/2)#

Now, #1/(1+t/2)# is the sum of a geometric series of ratio #(-t/2)#, so:

#1/(1-x) = -1/2 sum_(n=0)^oo (-t/2)^n#

and substituting back #-t = 3-x#:

#1/(1-x) = -1/2 sum_(n=0)^oo ((3-x)/2)^n#