Difference between revisions of "Sum of totient equals z/((1-z) squared)"

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(Created page with "==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 toti...")
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Revision as of 04:35, 22 June 2016

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