让我们从一个简单情况开始, 假设我们要计算这么一个常数:
$$
α=\sum _{k=0}^{∞}\frac{1}{10^k}\left(\frac{1}{k+3}+\frac{2}{k+5}\right)
$$
两边乘以 $10^d$ 并取小数, 则:
$$
\lbrace 10^d× α\rbrace = \left\{2×\sum _{k=0}^{\infty }
{\frac { 10^{d-k}}{k+1}}
\right\}
$$
$$\lbrace 10^{n} S_{p,q}\rbrace=\bigg \lbrace \bigg \lbrace q×\sum_{k=0}^{n}\frac{10^{n-k}}{k+p} \bigg \rbrace+q×\sum_{k=n+1}^{\infty}\frac{10^{n-k}}{k+p}\bigg \rbrace$$