Sum of sum of divisors function equals product of Riemann zeta for Re(z) greater than k+1

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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