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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==References== | ==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=}}: $24.3.2 I. | + | * {{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}}: $24.3.2 \mathrm{I}.B.$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 04:50, 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
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $24.3.2 \mathrm{I}.B.$
