已知 $n\in\mathbb N^*$，求证：$\ln \left(2^2+1\right)+\ln \left(3^2+1\right)+\cdots +\ln \left(n^2+1\right)<1+2\ln n!$．
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【解析】
欲证明$$\ln\left(2^2+1\right)+\ln\left(3^2+1\right)+\cdots+\ln\left(n^2+1\right)<1+2\ln n!,$$加强到$$\ln\left(2^2+1\right)+\ln\left(3^2+1\right)+\cdots+\ln\left(n^2+1\right)<1+2\ln n!-\dfrac 1{n+\dfrac 12}.$$只需要证明$$\ln \left(n^2+1\right)<2\ln n +\dfrac{1}{\left(n-\dfrac 12\right)\left(n+\dfrac 12\right)},$$即$$\ln\left(1+\dfrac{1}{n^2}\right)<\dfrac{1}{\left(n-\dfrac 12\right)\left(n+\dfrac 12\right)}.$$而我们熟知当 $x>0$ 时，有$$\ln(1+x)<\dfrac{2x}{x+2},$$于是$$\ln\left(1+\dfrac{1}{n^2}\right)<\dfrac{2}{2n^2+1}<\dfrac{1}{\left(n-\dfrac 12\right)\left(n+\dfrac 12\right)},$$命题得证．
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