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

so:

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

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

and clearly in general:

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

The coefficients of the Taylor series around #x=2# are then:

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

so that:

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

or:

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

We can also find the radius of convergence of this series by noting that:

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

We know that the sum of a geometric series is:

#sum_(n=0)^oo alpha^n = 1/(1-alpha)#

and is convergent in the interval #alpha in (-1,1)#

So we have:

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

converging for:

#-1 < 1-x/2 < 1#

#-2 < -x/2 < 0#

#0 < x/2 <= 2#

#0 < x < 4#