What is the Radius of Convergence for this power series? #ln(1-z) = - z - 1/2 z^2 - 1/3 z^3 ...#

Question

Answer key says: #abs z < 1#

And:
#z = x + iy#

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Answer 1 · 2017-03-16T09:09:50

#abs z < 1#

#d/(dz)(z-1/2z^2+1/3z^3+cdots +(-1)^(n+1)/n z^n+cdots) = sum_(k=0)^oo (-1)^kz^k#

but

#sum_(k=0)^oo (-1)^k z^k=lim_(n->oo)(z^n+1)/(z+1)#. Now considering #abs z < 1# we have

#sum_(k=0)^oo (-1)^k z^k=1/(1+z)# and

#int sum_(k=0)^oo (-1)^kz^k dz = log(1+z)#

now making the substitution #z->-z# we have

#-int sum_(k=0)^oo z^k dz =-sum_(k=1)^oo z^k/k = log(1-z)#

so it is convergent for #abs z < 1#