Sum of totient equals z/((1-z) squared)
From specialfunctionswiki
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 totient.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $24.3.2 \mathrm{I}.B.$
