级数方法解答的一道积分不等式题

百年云烟只过眼，不为繁华易素心

一道略火的题，由雷神解答：

证明\[\int_0^{2\left( {k + 1} \right)} {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} > \frac{k}{{1 + k}},\quad k = 0,1,2, \cdots .\]

证.注意到\[\int {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} = - {e^{ - x}}\sum\limits_{i = 0}^k {\frac{{{x^i}}}{i}} + C.\]

因此

\[\int_0^{2\left( {k + 1} \right)} {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} = - \left. {{e^{ - x}}\sum\limits_{i = 0}^k {\frac{{{x^i}}}{i}} } \right|_0^{2\left( {k + 1} \right)} = 1 - {e^{ - 2\left( {k + 1} \right)}}\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{i}} .\]

等价于证明\[1 - {e^{ - 2\left( {k + 1} \right)}}\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{i}} > \frac{k}{{1 + k}}.\]

即证

\[{e^{2k + 2}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} \Leftrightarrow \sum\limits_{i = 0}^\infty {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .\]

下面证明\[\sum\limits_{i = 0}^{2k + 1} {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .\]

等价于证明

\[\sum\limits_{i = k + 1}^{2k + 1} {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} = \sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^{i + k + 1}}}}{{\left( {i + k + 1} \right)!}}} > k\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .\]

只需证明\[\frac{{{{\left( {2k + 2} \right)}^{i + k + 1}}}}{{\left( {i + k + 1} \right)!}} > k\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}} \Leftrightarrow {\left( {2k + 2} \right)^{k + 1}} > k\frac{{\left( {i + k + 1} \right)!}}{{i!}}.\]

而

\[{\left( {2k + 2} \right)^{k + 1}} > k\frac{{\left( {2k + 1} \right)!}}{{k!}} = k\left( {k + 1} \right) \cdots \left( {2k + 1} \right) > k\frac{{\left( {i + k + 1} \right)!}}{{i!}}\]是显然的.