#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#
