\(\displaystyle\sum_{n=1}^{\infty}\frac{n^A}{B^n}=\displaystyle\sum_{n=1}^{A}\frac{StirlingS2\cdot n!\cdot B}{(B-1)^{n+1}}\)
譬如： B = 2A, A = 1, 2, 3, 4, 5, 6, 7, 8, 9, ......
\(\displaystyle\sum_{n=1}^{\infty}\frac{n^A}{(2A)^n}=\displaystyle\sum_{n=1}^{A}\frac{StirlingS2\cdot n!\cdot 2A}{(2A-1)^{n+1}}=2, \frac{20}{27}, \frac{366}{625}, \frac{10440}{16807}, \frac{142870}{177147}, \frac{23961756}{19487171}, \frac{1745610902}{815730721}, \frac{3556490576}{854296875}, \frac{17979203734218}{2015993900449}, \cdots\cdots\)
变化一下——把分母去掉。
\(\displaystyle\sum_{n=1}^{A}\frac{StirlingS2\cdot n!}{(2A-1)^{n-A}}\) = {1, 5, 61, 1305, 42861, 1996813, 124686493, 10002629745, 998844651901, ......}——OEIS没有这串数——当然 "2" 是可以换掉的。
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