Difference between revisions of "Series for log(z) for Re(z) greater than 1/2"

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==Theorem==
 
==Theorem==
 
The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$:
 
The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$:
$$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{(-1)^k}{k},$$
+
$$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{1}{k},$$
 
where $\log$ denotes the [[logarithm]].
 
where $\log$ denotes the [[logarithm]].
  
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==References==
 
==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Taylor series of log(1+z)|next=Series for log(z) for |z-1| less than 1}}: 4.1.25
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Taylor series of log(1+z)|next=Series for log(z) for absolute value of (z-1) less than 1}}: $4.1.25$
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 17:28, 27 June 2016

Theorem

The following formula holds for $\mathrm{Re}(z) \geq \dfrac{1}{2}$: $$\log(z) = -\displaystyle\sum_{k=1}^{\infty} \left(\dfrac{z-1}{z} \right)^k \dfrac{1}{k},$$ where $\log$ denotes the logarithm.

Proof

References