Letting #f(x)=ln(1-x)#, you could use the formula #f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+cdots# to get the answer above.

However, it's more interesting (fun?) to use the geometric series #1/(1-x)=1+x+x^2+x^3+x^4+cdots# and integrate it term by term, using the fact that #ln(1-x)=-int 1/(1-x)\ dx#, with #C=0# since #ln(1-0)=ln(1)=0#.

Doing this gives:

#ln(1-x)=-int (1+x+x^2+x^3+x^4+cdots)\ dx#

#=C-x-x^2/2-x^3/3-x^4/4-cdots#

#=-x-x^2/2-x^3/3-x^4/4-cdots#

Since #1/(1-x)=1+x+x^2+x^3+cdots# for #|x|<1#, this implies that the radius of convergence is 1.

However, an interesting thing happens at #x=-1# for the series #-x-x^2/2-x^3/3-x^4/4-cdots#. It ends up equaling #1-1/2+1/3-1/4+cdots#, which is the so-called Alternating Harmonic Series, which converges (though not "absolutely"). Moreover, it also happens to equal #ln(1-(-1))=ln(2)#.

Hence, even though the radius of convergence is #1#, the series for #ln(1-x)# converges and equals #ln(1-x)# over the half-open/half-closed interval #[-1,1)# (it doesn't converge at #x=1# since it's the opposite of the Harmonic Series there).