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## 1.5 The Logarithm

### 1.51 Series representation

1.51112

$\mathrm{ln}\left(1+x\right)=x-\frac{1}{2}{x}^{2}+\frac{1}{3}{x}^{3}-\frac{1}{4}{x}^{4}+\dots =\sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}\frac{{x}^{k}}{k}=xF\left(1,1;\text{?}2;-x\right)$

$[−1

1.512

1.

$\mathrm{ln}x=\left(x-1\right)-\frac{1}{2}{\left(x-1\right)}^{2}+\frac{1}{3}{\left(x-1\right)}^{3}-\cdots =\sum _{k=1}^{\infty }{\left(-1\right)}^{k+1}\frac{{\left(x-1\right)}^{k}}{k}$

$[0

2.

$\mathrm{ln}x=2\left[\frac{x-1}{x+1}+\frac{1}{3}{\left(\frac{x-1}{x+1}\right)}^{3}+\frac{1}{5}{\left(\frac{x-1}{x+1}\right)}^{5}+\cdots \right]=2\sum _{k=1}^{\infty }\frac{1}{2k-1}{\left(\frac{x-1}{x+1}\right)}^{2k-1}$

$[0

3.

$\mathrm{ln}x=\frac{x-1}{x}+\frac{1}{2}{\left(\frac{x-1}{x}\right)}^{2}+\frac{1}{3}{\left(\frac{x-1}{x}\right)}^{3}+\cdots =\sum _{k=1}^{\infty }\frac{1}{k}{\left(\frac{x-1}{x}\right)}^{k}$

AD (644.6)

$[x≥12]$

4.*

$\mathrm{ln}$

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