用数学归纳法证明\(\left( {1}+\dfrac{{1}}{{3}} \right)\left( {1}+\dfrac{{1}}{{5}} \right)\left( {1}+\dfrac{{1}}{{7}} \right)\ldots \left( {1}+\dfrac{{1}}{{2}n-{1}} \right) > \dfrac{\sqrt{{2}n+{1}}}{{2}}(n∈N^{*}\)且\(n\geqslant 2)\)

已知数列\(\{ a_{n}\}\)满足\(a_{1}{=}1\)，\(a_{n{+}1}{=}\dfrac{a_{n}}{1{+}{a_{n}}^{2}}\)，\(a_{n{+}1}{=}\dfrac{a_{n}}{1{+}{a_{n}}^{2}}.\)记\(S_{n}\)，\(T_{n}\)分别是数列\(\{ a_{n}\}\)，\(\{{a_{n}}^{2}\}\)的前\(n\)项和，证明：当\(n{∈}\mathbf{N}^{\mathbf{{*}}}\)时，

\((1)a_{n{+}1}{ < }a_{n}\)；

\((2)T_{n}{=}\dfrac{1}{{a_{n{+}1}}^{2}}{-}2n{-}1\)；

\((3)\sqrt{2n}{-}1{ < }S_{n}{ < }\sqrt{2n}\)．

已知数列\(\left\{ {{a}_{n}} \right\}\)中，\({{a}_{1}}=1\)且\({{a}_{n+1}}={{a}_{n}}+2n+1\)，设数列\(\left\{ {{b}_{n}} \right\}\)满足\({{b}_{n}}={{a}_{n}}-1\)，对任意正整数\(n\)，不等式\(\dfrac{1}{{{b}_{2}}}+\dfrac{1}{{{b}_{3}}}+\bullet \bullet \bullet +\dfrac{1}{{{b}_{n}}} < m\)均成立，则实数\(m\)的取值范围为           ．

已知数列\(\left\{ {{a}_{n}} \right\}\)满足\({{a}_{1}}=1\)，\({{a}_{n+1}}=\dfrac{{{a}_{n}}}{1+a_{n}^{2}}\)，\(n\in {{N}^{*}}\)，记\({{S}_{n}}\)，\({{T}_{n}}\)分别是数列\(\left\{ {{a}_{n}} \right\}\)，\(\left\{ a_{n}^{2} \right\}\)的前\(n\)项和，证明：当\(n\in {{N}^{*}}\)时，\((1)\)\({{a}_{n+1}} < {{a}_{n}}\)；\((2)\)\({{T}_{n}}=\dfrac{1}{a_{n+1}^{2}}-2n-1\)；\((3)\)\(\sqrt{2n}-1 < {{S}_{n}} < \sqrt{2n}\)．

已知函数\(f(x)\) \(=x-1-a\ln x\)．

\((1)\)若\(a=2\)，求\(y=f(x)\)在\((4,f(4))\)处的切线方程

\((2)\)就\(a\in R\)讨论，\(f(x)\)的最小值，并求\(f(x)\geqslant 0\)时\(a\)的值；

\((3)\)设\(m\)为整数，且对于任意正整数\(n\)，\((1{+}\dfrac{1}{2})(1{+}\dfrac{1}{{{2}^{2}}})\ldots (1{+}\dfrac{1}{{{2}^{n}}})﹤m\)，求\(m\)的最小值．

设\(S\)\({\,\!}_{n}\)是数列\(\{\)\(a_{n}\)\(\}\)的前\(n\)项和，\(a_{n}\)\( > 0\)，且\(4S\)\({\,\!}_{n}\)\(=\)\(a_{n}\)\((\)\(a_{n}\)\(+2)\)．

\((\)Ⅱ\()\)设\(b_{n}\)\(= \dfrac{1}{\left({a}_{n}-1\right)\left({a}_{n}\;+1\right)} \)，\(T\)\({\,\!}_{n}\)\(=\)\(b\)\({\,\!}_{1}+\)\(b\)\({\,\!}_{2}+…+\)\(b_{n}\)，求证：\(T\)\({\,\!}_{n}\)\( < \dfrac{1}{2} \)．

已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，\(2S_{n}=3a_{n}-2n(n∈N^{+}).\)

\((1)\)证明数列\(\{a_{n}+1\}\)是等比数列，并求数列\(\{a_{n}\}\)的通项公式；

\((2)\)设\(b_{n}=a_{n}+2^{n}+1\)，求证：\( \dfrac{1}{{b}_{1}}+ \dfrac{1}{{b}_{2}}+...+ \dfrac{1}{{b}_{n}} < \dfrac{1}{2}- \dfrac{1}{{2}^{n+1}} \)．

已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\)，且满足\(a_{n}+2S_{n}·S_{n-1}=0(n\geqslant 2)\)，\(a_{1}= \dfrac{1}{2}\)．

\((1)\)求证：\(\{ \dfrac{1}{S_{n}}\}\)是等差数列；

\((2)\)求\(a_{n}\)的表达式；

\((3)\)若\(b_{n}=2(1-n)a_{n}(n\geqslant 2)\)，求证：\(b\rlap{_{2}}{^{2}}+b\rlap{_{3}}{^{2}}+…+b\rlap{_{n}}{^{2}} < 1\)．