Derivative of Riemann Zeta Function

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Theorem

The derivative of the Riemann zeta function is:

$\displaystyle \zeta' \left({z}\right) = \frac{\mathrm d \zeta}{\mathrm dz} = -\sum_{n \mathop = 2}^\infty \frac{\ln \left({n}\right)}{n^z}$


Proof

\(\displaystyle \frac{\mathrm d \zeta}{\mathrm dz}\) \(=\) \(\displaystyle \frac{\mathrm d}{\mathrm dz} \left({\sum_{n \mathop = 1}^\infty n^{-z} }\right)\)
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mathrm d}{\mathrm dz} \left({n^{-z} }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left({- \ln \left({n}\right) n^{-z} }\right)\) from Derivative of Exponential Function
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 1}^\infty \frac{\ln \left({n}\right)}{n^z}\)
\(\displaystyle \) \(=\) \(\displaystyle -\sum_{n \mathop = 2}^\infty \frac{\ln \left({n}\right)}{n^z}\) as $\ln 1 = 0$

$\blacksquare$