\(a^{-k}\)及び\(a^{-kx}\)の無限級数
\(a^{-k}\)の無限級数
$$\begin{eqnarray}
S=\sum_{k=0}^{\infty}a^{-k}
&=&\sum_{k=0}^{\infty}\left(a^{-1}\right)^k
\;\ldots\;a\in\mathbb{R},\;\left|a\right|^{-1}=\frac{1}{\left|a\right|}\lt1
\\&=&\sum_{k=0}^{\infty}\left(\frac{1}{a}\right)^k
\\&=&\sum_{k=0}^{\infty}\frac{1}{a^{k}}
\\&=&\frac{1}{a^0}+\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}+\cdots
\\&=&1+\frac{1}{a}+\frac{1}{a\cdot a}+\frac{1}{a\cdot a^{2}}+\cdots
\\&=&1+\frac{1}{a}+\frac{1}{a}\frac{1}{a}+\frac{1}{a}\frac{1}{a^{2}}+\cdots
\\&=&1+\frac{1}{a}\left(1+\frac{1}{a}+\frac{1}{a^{2}}+\cdots\right)
\\&=&1+\frac{1}{a}S
\\S-\frac{1}{a}S&=&1
\\S\left(1-\frac{1}{a}\right)&=&1
\\S&=&\frac{1}{1-\frac{1}{a}}
\\&=&\frac{1}{\frac{a-1}{a}}
\\&=&\frac{a}{a-1}
\end{eqnarray}$$
\(a^{-kx}\)の無限級数
$$\begin{eqnarray}
S=\sum_{k=0}^{\infty}a^{-kx}&=&\sum_{k=0}^{\infty}\left(a^{-x}\right)^k
\;\ldots\;a,x\in\mathbb{R},\;|a|^{-x}=\frac{1}{|a|^{x}}\lt1
\\&=&\sum_{k=0}^{\infty}\left(\frac{1}{a^{x}}\right)^k
\\&=&\sum_{k=0}^{\infty}\frac{1}{a^{kx}}
\\&=&\frac{1}{a^{0x}}+\frac{1}{a^x}+\frac{1}{a^{2x}}+\frac{1}{a^{3x}}+\cdots
\\&=&1+\frac{1}{a^{x}}+\frac{1}{a^{x}a^{x}}+\frac{1}{a^{x}a^{2x}}+\cdots
\\&=&1+\frac{1}{a^{x}}+\frac{1}{a^{x}}\frac{1}{a^{x}}+\frac{1}{a^{x}}\frac{1}{a^{2x}}+\cdots
\\&=&1+\frac{1}{a^{x}}\left(1+\frac{1}{a^{x}}+\frac{1}{a^{2x}}+\cdots\right)
\\&=&1+\frac{1}{a^{x}}S
\\S-\frac{1}{a^x}S&=&1
\\S\left(1-\frac{1}{a^x}\right)&=&1
\\S&=&\frac{1}{1-\frac{1}{a^x}}
\\&=&\frac{1}{\frac{a^x-1}{a^x}}
\\&=&\frac{a^x}{a^x-1}
\end{eqnarray}$$
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