a^(-k)及びa^(-kx)の無限級数

$a^{-k}$及び$a^{-kx}$の無限級数

$a^{-k}$の無限級数

$$\begin{array}{rcl}S=\sum _{k=0}^{\mathrm{\infty }}{a}^{-k}& =& \sum _{k=0}^{\mathrm{\infty }}{\left(a^{-1}\right)}^k\dots a\in \mathbb{R},{\left|a\right|}^{-1}=\frac{1}{\left|a\right|}<1\\ & =& \sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{a}\right)}^k\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{a^k}\\ & =& \frac{1}{a^0}+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}+\cdots \\ & =& 1+\frac{1}{a}+\frac{1}{a\cdot a}+\frac{1}{a\cdot a^2}+\cdots \\ & =& 1+\frac{1}{a}+\frac{1}{a}\frac{1}{a}+\frac{1}{a}\frac{1}{a^2}+\cdots \\ & =& 1+\frac{1}{a}(1+\frac{1}{a}+\frac{1}{a^2}+\cdots )\\ & =& 1+\frac{1}{a}S\\ S-\frac{1}{a}S& =& 1\\ S(1-\frac{1}{a})& =& 1\\ S& =& \frac{1}{1-\frac{1}{a}}\\ & =& \frac{1}{\frac{a-1}{a}}\\ & =& \frac{a}{a-1}\end{array}$$

$a^{-kx}$の無限級数

$$\begin{array}{rcl}S=\sum _{k=0}^{\mathrm{\infty }}{a}^{-kx}& =& \sum _{k=0}^{\mathrm{\infty }}{\left(a^{-x}\right)}^k\dots a,x\in \mathbb{R},|a{|}^{-x}=\frac{1}{|a{|}^x}<1\\ & =& \sum _{k=0}^{\mathrm{\infty }}{\left(\frac{1}{a^x}\right)}^k\\ & =& \sum _{k=0}^{\mathrm{\infty }}\frac{1}{a^{kx}}\\ & =& \frac{1}{a^{0x}}+\frac{1}{a^x}+\frac{1}{a^{2x}}+\frac{1}{a^{3x}}+\cdots \\ & =& 1+\frac{1}{a^x}+\frac{1}{a^x{a}^x}+\frac{1}{a^x{a}^{2x}}+\cdots \\ & =& 1+\frac{1}{a^x}+\frac{1}{a^x}\frac{1}{a^x}+\frac{1}{a^x}\frac{1}{a^{2x}}+\cdots \\ & =& 1+\frac{1}{a^x}(1+\frac{1}{a^x}+\frac{1}{a^{2x}}+\cdots )\\ & =& 1+\frac{1}{a^x}S\\ S-\frac{1}{a^x}S& =& 1\\ S(1-\frac{1}{a^x})& =& 1\\ S& =& \frac{1}{1-\frac{1}{a^x}}\\ & =& \frac{1}{\frac{a^x-1}{a^x}}\\ & =& \frac{a^x}{a^x-1}\end{array}$$