已知曲线\(f(x)= \dfrac{{\log }_{2}(x+1)}{x+1}(x > 0) \)上有一点列\({P}_{n}({x}_{n},{y}_{n})(n∈{N}_{∗}) \)，过点\({P}_{n} \)在\(x \)轴上的射影是\({Q}_{n}({x}_{n},0) \)，且\({x}_{1}+{x}_{2}+{x}_{3}+⋯{x}_{n}={2}^{n+1}−n−2. (n∈{N}_{∗}) \)

\((1)\)求数列\(\{{x}_{n}\} \)的通项公式

\((2)\)设四边形\({P}_{n}{Q}_{n}{Q}_{n+1}{P}_{n+1} \)的面积是\({S}_{n} \)，求\({S}_{n} \)

\((3)\)在\((2) \)条件下，求证：\(\dfrac{1}{{S}_{1}}+ \dfrac{1}{2{S}_{2}}+⋯+ \dfrac{1}{n{S}_{n}} < 4. \)

曲线\(y=\sqrt{x}\)上点\({{P}_{1}},{{P}_{2}},\cdots {{P}_{n}}\)与\(x\)轴上点\({{Q}_{1}},{{Q}_{2}},\cdots {{Q}_{n}}\)构成一列正三角形，即\(\Delta O{{P}_{1}}{{Q}_{1}},\Delta {{Q}_{1}}{{P}_{2}}{{Q}_{2}},\cdots \Delta {{Q}_{n-1}}{{P}_{n}}{{Q}_{n}}\)，设\(\Delta O{{P}_{1}}{{Q}_{1}}\)的边长为\({{a}_{1}}\)，正三角形\(\Delta {{Q}_{n-1}}{{P}_{n}}{{Q}_{n}}\)的边长为\({{a}_{n}}\)

\((1)\)求\({{a}_{1}},{{a}_{2}}\)

\((2)\)求数列\(\left\{ {{a}_{n}} \right\}\)的通项公式

\((3)\)证明：\(\dfrac{1}{{{a}_{1}}^{2}}+\dfrac{1}{{{a}_{2}}^{2}}+\cdots +\dfrac{1}{{{a}_{n}}^{2}} < \dfrac{63}{16}\)

已知曲线\(f(x)=\dfrac{{\log }_{2}\left(x+1\right)}{x+1} (x > 0)\)上有一点列\(P_{n}(x_{n},y_{n})(n∈N*)\)，过点\(P_{n}\)在\(x\)轴上的射影是\(Q_{n}(x_{n},0)\)，且\(x_{1}+x_{2}+x_{3}+…+x_{n}=2^{n+1}-n-2.(n∈N*)\)

\((1)\)求数列\(\{x_{n}\}\)的通项公式；

\((2)\)设四边形\(P_{n}Q_{n}Q_{n+1}P_{n+1}\)的面积是\(S_{n}\)，求\(S_{n}\)；

\((3)\)在\((2)\)条件下，求证：\(\dfrac{1}{{S}_{1}} \) \(+\dfrac{1}{2{S}_{2}} +…+\dfrac{1}{n{S}_{n}} < 4\)．

设函数\(f\)\((\)\(x\)\()=\dfrac{1-x}{ax}+\ln x\)在\([1,+∞)\)上为增函数\(.\)  

\((1)\)求正实数\(a\)的取值范围；

\((2)\)比较\( \dfrac{1}{n} \)与\(\ln \dfrac{n}{n-1}\left(n∈{N}^{*},n\geqslant 2\right) \)的大小，说明理由；

\((3)\)求证：\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots +\dfrac{1}{n} < \ln n(\)\(n∈N\)\(*\)， \(n\)\(\geqslant 2)\)