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本帖最后由 王守恩 于 2020-11-19 20:12 编辑
elim 发表于 2020-11-18 16:12
\(\displaystyle\small\sum_{n=1}^{\infty}\frac{a^n}{(a+1)^n}=\frac{a}{a+1}\frac{a+1}{a+1-a}=a\;(a>0)\ ...
从简单说起。
Table[\(\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{(a+1)^n}\), {a, 1, 25}]
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}
Table[\(\displaystyle\sum_{n=1}^{\infty}\frac{a^{n+1}}{(a+1)^n}\), {a, 1, 24}]
{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576}
Table[\(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(a+1)^n}\), {a, 1, 15}]
{1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12, 1/13, 1/14, 1/15}
Table[\(\displaystyle\sum_{n=1}^{\infty}\frac{a^n}{(2a+1)^n}\), {a, 1, 15}]
{1/2, 2/3, 3/4, 4/5, 5/6, 6/7, 7/8, 8/9, 9/10, 10/11, 11/12, 12/13, 13/14, 14/15, 15/16} |
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