Difference between revisions of "Series for log(z) for Re(z) greater than 1/2"
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(Created page with "==Theorem== The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$: $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{(-1)^k}{k},...") |
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Revision as of 07:32, 4 June 2016
Theorem
The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$: $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{(-1)^k}{k},$$ where $\log$ denotes the logarithm.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous): 4.1.25
