已知数列 $\{a_n\}$ 的通项 $a_n=\sqrt{\dfrac{n}{n+2}}-\dfrac{n}{n+1}$．
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【标注】
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求证：数列 $\{a_n\}$ 单调递减；标注答案略解析只需要证明\[a_{n+1}<a_n,\]即\[\sqrt{\dfrac{n+1}{n+3}}-\dfrac{n+1}{n+2}<\sqrt{\dfrac{n}{n+2}}-\dfrac n{n+1},\]也即\[\sqrt{\dfrac{n+1}{n+3}}-\sqrt{\dfrac{n}{n+2}}<\dfrac{n+1}{n+2}-\dfrac {n}{n+1},\]也即\[\dfrac{\dfrac{n+1}{n+3}-\dfrac{n}{n+2}}{\sqrt{\dfrac{n+1}{n+3}}+\sqrt{\dfrac{n}{n+2}}}<\dfrac{n+1}{n+2}-\dfrac{n}{n+1},\]也即\[\dfrac{\dfrac{2}{(n+2)(n+3)}}{\dfrac{1}{(n+1)(n+2)}}<\sqrt{\dfrac{n+1}{n+3}}+\sqrt{\dfrac n{n+2}},\]也即\[\dfrac{2(n+1)}{n+3}<\sqrt{\dfrac{n+1}{n+3}}+\sqrt{\dfrac{n}{n+2}}.\]事实上，有\[\dfrac{n+1}{n+3}<\sqrt{\dfrac{n+1}{n+3}},\dfrac{n+1}{n+3}<\sqrt{\dfrac{n}{n+2}},\]于是原命题得证．
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若 $S_n$ 是数列 $\{a_n\}$ 的前 $n$ 项和，求证：$S_n<\dfrac 14$．标注答案略解析考虑裂项放缩，有\[\begin{split} a_n&=\sqrt{\dfrac{n}{n+2}}-\dfrac{n}{n+1}\\
&=\dfrac{(n+1)\sqrt n-n\sqrt{n+2}}{(n+1)\sqrt{n+2}}\\
&=\dfrac{\sqrt{n}}{(n+1)\sqrt{n+2}\cdot \left(\sqrt{n^2+2n+1}+\sqrt{n^2+2n}\right)}\\
&<\dfrac{1}{2(n+1)(n+2)}\\
&=\dfrac {1}{2(n+1)}-\dfrac{1}{2(n+2)},\end{split}\]于是\[S_n<\dfrac 14-\dfrac{1}{2(n+2)}<\dfrac 14,\]原命题得证．
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