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好像是老题，懒得找链接了，写一下也蛮简单
\begin{align*}
a_n&=\sum_{k=1}^n\frac1{k(n+1-k)} \\
& =\sum_{k=1}^n\frac1{n+1}\left( \frac1k+\frac1{n+1-k} \right) \\
&=\frac1{n+1}\left( \sum_{k=1}^n\frac1k+\sum_{k=1}^n\frac1{n+1-k} \right) \\
&=\frac2{n+1}\sum_{k=1}^n\frac1k,
\end{align*}
故
\begin{align*}
a_n-a_{n+1}&=\frac2{n+1}\sum_{k=1}^n\frac1k-\frac2{n+2}\sum_{k=1}^{n+1}\frac1k \\
&=\frac2{n+1}\sum_{k=1}^n\frac1k-\frac2{n+2}\sum_{k=1}^n\frac1k-\frac2{(n+1)(n+2)} \\
&=\frac2{(n+1)(n+2)}\sum_{k=1}^n\frac1k-\frac2{(n+1)(n+2)} \\
&=\frac2{(n+1)(n+2)}\sum_{k=2}^n\frac1k\\
&>0.
\end{align*} |
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kuing
发表于 2014-4-8 15:25
