已知函数\(f(x)=(x+2)\ln x,g(x)=x^2+(3-a)x+2(1-a)\)
(1)若不等式\(f(x)\leq g(x)\)在\(x\in(-2,+\infty)\)上恒成立，求\(a\)取值范围.
(2)证明:\(\displaystyle \sum\limits_{k=1}^{n}\left(1+\dfrac{1}{4^k}\right)<e^{\frac{1}{3}}\)
(1) \(f(x)\leq g(x)\) 转化为
\[a\leq (x+1-\ln(x+2))_{\min}
\]
记\(h(x)=x+1-\ln(x+2)\),\(h^{\prime}(x)=\dfrac{x+1}{x+2}\)
不难得到\(h(x)_{\min}=h(1)=0\)
从而\(a\leq 0\)
\[\ln\left(1+\dfrac{1}{4}\right)\leq \dfrac{1}{4}
\]
\[\ln\left(1+\dfrac{1}{4^2}\right)\leq \dfrac{1}{4^2}
\]
\[\vdots
\]
\[\ln\left(1+\dfrac{1}{4^n}\right)\leq \dfrac{1}{4^n}
\]
累加有
\[\ln\prod\limits_{k=1}^{n}\left(1+\dfrac{1}{4^k}\right)=\sum\limits_{k=1}^{n}\dfrac{1}{4^k}=\dfrac{1}{3}\left(1-\dfrac{1}{4^n}\right)<\dfrac{1}{3}
\]
则
\[\prod\limits_{k=1}^{n}\left(1+\dfrac{1}{4^k}\right)<e^{\frac{1}{3}}
\]