firstly we look at the formula for the Taylor series, which is:
#f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n #
which equals:
#f(a) + f'(a)(x-a) + (f''(a)(x-a)^2)/(2!) + (f'''(a)(x-a)^3)/(3!) + ... #
So you would like to solve for #f(x) = ln(x)# at #x=1# which I assume mean centered at #1# of which you would make #a=1#
To solve:
#f(x) = ln(x)# and #f(1) = ln(1) = 0#
#f'(x) = 1/x# and #f'(1) = 1/1 = 1#
#f''(x) = -1/x^2# and #f''(1) = -1/(1)^2 = -1#
#f^((3))(x) = 2/x^3# and #f^((3))(1) = 2/(1)^3 = 2#
#f^((4))(x) = -((2)(3))/x^4# and #f^((4))(1) = -((2)(3))/(1)^4 = -(2)(3)#
Where now we can already start to see a pattern forming, so we starting using our formula(2):
#0 + 1(x-1) - (1(x-1)^2)/(2!) + (2(x-1)^3)/(3!) - ((2)(3)(x-1)^4)/(4!) .....#
and now try try see how we can write this as a series, which we get: (we start will n=1 as our first term is 0)
#f(x) = ln(x) = sum_(n=1)^oo (-1)^(n-1) (((n-1)!)(x-1)^n)/(n!) #
Which can then simplify to:
#f(x) = ln(x) = sum_(n=1)^oo (-1)^(n-1) (x-1)^n/n#
