View source for Sum of totient equals z/((1-z) squared)

==Theorem==
The following formula holds for $|z|<1$:
$$\displaystyle\sum_{k=1}^{\infty} \dfrac{\phi(k)x^k}{1-x^k}= \dfrac{x}{(1-x)^2} ,$$
where $\phi$ denotes the [[Euler totient|totient]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Sum of totient equals zeta(z-1)/zeta(z) for Re(z) greater than 2|next=Product representation of totient}}: $24.3.2 I.B.$

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