翻看了一下坛子里的相关内容，觉得应该是发在这里最合适了。要是这里不合适，请大家告知我一下。
我在对一化学过程数学建模时，得到一数学关系，是一数列，其中各项表达如下：
[latex]{x_1} = \frac{1}{{1 + \frac{b}{a} \cdot {1^{ - \frac{1}{2}}}}} \cdot {x_0}[/latex]
[latex]{x_2} = \frac{{{{\left( {\frac{1}{2}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {2^{ - \frac{7}{6}}}}} \cdot {x_1}[/latex]
[latex]{x_3} = \frac{{{{\left( {\frac{2}{3}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {3^{ - \frac{7}{6}}}}} \cdot {x_2}[/latex]
[latex]\vdots[/latex]
[latex]{x_i} = \frac{{{{\left( {\frac{{i - 1}}{i}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {i^{ - \frac{7}{6}}}}} \cdot {x_{i - 1}}[/latex]
其中：i = 自然数，即1, 2, 3, ... 正无穷
a, b = 均为不同的、大于0的正整数
[latex]{x_0}[/latex]= 大于0的正整数
求：在i趋于无穷大时，此数列的和，即：[latex]\sum\limits_{i = 1}^\infty{{x_i}}[/latex]
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目前，我自己做了下面的尝试，把此数列中各项都用[latex]{x_0}[/latex]来表达，即：
[latex]{x_1} = \frac{1}{{1 + \frac{b}{a} \cdot {1^{ - \frac{1}{2}}}}} \cdot {x_0} = \frac{1}{{1 + \frac{b}{a} \cdot {1^{ - \frac{7}{6}}}}} \cdot {x_0}[/latex]
[latex]{x_2} = \frac{{{{\left( {\frac{1}{2}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {2^{ - \frac{7}{6}}}}} \cdot {x_1} = \frac{{{{\left( {\frac{1}{2}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {2^{ - \frac{7}{6}}}}} \cdot \frac{1}{{1 + \frac{b}{a} \cdot {1^{ - \frac{7}{6}}}}} \cdot {x_0}[/latex]
[latex]{x_3} = \frac{{{{\left( {\frac{2}{3}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {3^{ - \frac{7}{6}}}}} \cdot {x_2} = \frac{{{{\left( {\frac{2}{3}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {3^{ - \frac{7}{6}}}}} \cdot \frac{{{{\left( {\frac{1}{2}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {2^{ - \frac{7}{6}}}}} \cdot \frac{1}{{1 + \frac{b}{a} \cdot {1^{ - \frac{7}{6}}}}} \cdot {x_0}[/latex]
[latex]\vdots[/latex]
[latex]{x_i} = \frac{{{{\left( {\frac{{i - 1}}{i}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {i^{ - \frac{7}{6}}}}} \cdot {x_{i - 1}} = \frac{{{{\left( {\frac{{i - 1}}{i}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {i^{ - \frac{7}{6}}}}} \cdot \frac{{{{\left( {\frac{{i - 2}}{{i - 1}}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {{\left( {i - 1} \right)}^{ - \frac{7}{6}}}}} \cdot \frac{{{{\left( {\frac{{i - 3}}{{i - 2}}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {{\left( {i - 2} \right)}^{ - \frac{7}{6}}}}} \cdot\cdots\cdot \frac{{{{\left( {\frac{2}{3}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {3^{ - \frac{7}{6}}}}} \cdot \frac{{{{\left( {\frac{1}{2}} \right)}^{\frac{2}{3}}}}}{{1 + \frac{b}{a} \cdot {2^{ - \frac{7}{6}}}}} \cdot \frac{1}{{1 + \frac{b}{a} \cdot {1^{ - \frac{7}{6}}}}} \cdot {x_0} = \frac{{\left[ {{{\left( {\frac{{i - 1}}{i}} \right)}^{\frac{2}{3}}}} \right]!}}{{\left( {1 + \frac{b}{a} \cdot {i^{ - \frac{7}{6}}}} \right)!}} \cdot {x_0}[/latex]
如果这样没错的话，则得到[latex]{x_i}[/latex]的通项公式为一阶乘表达式(我不确定，这个用阶乘表达的通项公式我是否写的数学上正确)。但到了这步后，我就进行不下去了，不知该怎样求此数列的和。google了一下，似乎有提到Stirling's approximation，不知对口不对口。
恳请大家的帮助！非常感谢先！
[Last edited by xc79522 on -8-30 at 11:23]
