Difference between revisions of "Sum of sum of divisors function equals product of Riemann zeta for Re(z) greater than k+1"
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(Created page with "__NOTOC__ ==Theorem== The following formula holds for $\mathrm{Re}(z)>k+1$: $$\displaystyle\sum_{k=1}^{\infty} \dfrac{\sigma_n(k)}{k^z} = \zeta(z) \zeta(z-n),$$ where $\sigma_...") |
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==References== | ==References== | ||
| − | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Sum of divisors|next=}}: $24.3.3 I.B.$ | + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Sum of divisors|next=findme}}: $24.3.3 I.B.$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] | ||
Latest revision as of 22:18, 25 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z)>k+1$: $$\displaystyle\sum_{k=1}^{\infty} \dfrac{\sigma_n(k)}{k^z} = \zeta(z) \zeta(z-n),$$ where $\sigma_n$ denotes the sum of divisors function and $\zeta$ denotes the Riemann zeta.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $24.3.3 I.B.$
