User:Ciantic/Sandbox

$$\frac{1}{e^x \pm 1} = \sum_{n=0}^\infty (-1)^{n} e^{-(n+1)x} = \sum_{n=1}^\infty(-1)^{n+1} e^{-n \cdot x}$$

Nämä pitänee paikkansa:

$$\frac{1}{1+x} = \sum_{n=0}^\infty (-1)^n x^n \quad\mbox{ for } |x|<1$$

$$\frac{1}{1-x} = \sum_{i=0}^\infty x^i \quad\mbox{ for } |x|\le 1, x\neq 1$$, ja

siis

$$\frac{1}{e^x + 1} = \frac{1}{1 + e^x} = \sum_{n=0}^\infty (-1)^n e^{x \cdot n} \neq \sum_{n=1}^\infty (-1)^{n+1} e^{-n \cdot x} \quad\mbox{ for } |e^x|<1$$, ja

$$\frac{1}{e^x - 1} = \frac{1}{-(1 - e^x)} = -\frac{1}{1 - e^x} = -\sum_{n=0}^\infty e^{x \cdot n} \quad\mbox{ for } |e^x|\le 1, e^x\neq 1$$