View source for Taylor series of log(1-z)

==Theorem==
The following formula holds:
$$\log(1-z)=-\displaystyle\sum_{k=1}^{\infty} \dfrac{z^k}{k},$$
where $\log$ denotes the [[logarithm]].

==Proof==

=References=
* {{BookReference|Dilogarithms and Associated Functions|1958|Leonard Lewin|prev=Dilogarithm|next=Relationship between dilogarithm and log(1-z)/z}}: $(1.2)$
*{{BookReference|Polylogarithms and Associated Functions|1981|ed=2nd|edpage=Second Edition|Leonard Lewin|prev=Dilogarithm|next=Relationship between dilogarithm and log(1-z)/z}}: $(1.2)$

[[Category:Theorem]]
[[Category:Unproven]]