$例1、已知数列\{a_n\}中，a_1=2,a_{n+1}=a_n+n+1.$
$(1)求数列\{a_n\}的通项公式；$
$(2)设b_n=\cfrac{1}{a_n},数列\{b_n\}的前n项和为T_n，证明T_n\lt 2$

$(1)a_{n+1}=a_n+n+1，有递推关系式：$
$a_{n}=a_{n-1}+(n-1)+1$
$a_{n-1}=a_{n-2}+(n-2)+1$
$a_{n-2}=a_{n-3}+(n-3)+1$
$a_{n-3}=a_{n-4}+(n-4)+1$
$\cdots \cdots \cdots\cdots$
$a_3=a_2+2+1$
$a_2=a_1+1+1$
累加之
$a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\cdots +a_3+a_2=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+\cdots +a_2+a_1$
$+(n-1)+(n-2)+(n-3)+(n-4)+\cdots +1+1+1\cdots +1$
$a_n=a_1+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1(n-1)\times 1$
$=2+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1+(n-1)$
$=n+(n-1)+\cdots +1+1=\cfrac{n(n+1)}{2}+1=\cfrac{n^2+n+2}{2}$
$b_n=\cfrac{1}{a_n}=\cfrac{2}{n^2+n+2}\lt {\color{Red} \cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})} $
$b_1\lt 2(1-\cfrac{1}{2})$
$b_2\lt 2(\cfrac{1}{2}-\cfrac{1}{3})$
$b_3\lt 2(\cfrac{1}{3}-\cfrac{1}{4})$
$\cdots\cdots\cdots\cdots$
$b_{n-2}\lt 2(\cfrac{1}{n-2}-\cfrac{1}{n-1})$
$b_{n-1}\lt 2(\cfrac{1}{n-1}-\cfrac{1}{n})$
$b_n\lt 2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$T_n\lt 2(1-\cfrac{1}{n+1})\lt 2$

$例2、已知数列\{a_n\}的前n项和为S_n,且a_2=3,a_{n+1}=S_n+n+1$
$(1)证明数列\{a_n+1\}是等比数列$
$递推公式：a_{n+1}=S_n+n+1\quad ①\Rightarrow a_2=S_1+1+1=a_1+2,a_1=1$
$\qquad\qquad a_{n}=S_{n-1}+(n-1)+1\quad ②$
$①-②,得a_{n+1}－a_{n}＝S_n+n+1－[S_{n-1}+(n-1)+1]$
$a_{n+1}－a_{n}=a_n+1\Rightarrow a_{n+1}+1=2(a_n+1)$
$故数列\{a_n+1\}是以a_1+1=2为首项，q=2的等比数列$

$15、d=\cfrac{S_2}{2}-\cfrac{S_1}{1}=\cfrac{a_2+a_1}{2}-\cfrac{a_1}{1}=\cfrac{3-a_1}{2}$
$\cfrac{S_5}{5}=S_1+4d=a_1+4d=a_1+4\times \cfrac{3-a_1}{2}=6-a_1=5\Rightarrow a_1=1,d=1$
$\cfrac{S_n}{n}=n\Rightarrow S_n=n^2\Rightarrow n\ge 2,S_{n-1}=(n-1)^2,S_n-S_{n-1}=a_n=2n-1(n\ge 2)$
$n=1时，a_1=1,故a_n=2n-1在n=1时也成立。$
$(2)、b_n=2(n+1)(a_n+1)=4n(n+1)\Rightarrow \cfrac{1}{b_n}=\cfrac{1}{4n(n+1)}=\cfrac{1}{4}(\cfrac{1}{n}-\cfrac{1}{n+1})$
$\cfrac{1}{b_1}+\cfrac{1}{b_2} +\cdots +\cfrac{1}{b_n}=\cfrac{1}{4}(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cfrac{1}{3}-\cfrac{1}{4}+\cdots+\cfrac{1}{n-1}-\cfrac{1}{n}+\cfrac{1}{n}-\cfrac{1}{n+1})$
$=\cfrac{1}{4}(1-\cfrac{1}{n+1})\lt \cfrac{1}{4}$

26年江苏省高三G4联考第17题
$已知数列\{a_n\}满足a_1=1,\cfrac{S_{n+1}}{S_n}=\cfrac{n+2}{n}$
$(1)求数列\{a_n\}的通项公式；$
$(2)设b_n=a_n^2(\cos^2\cfrac{n\pi}{3}-\sin^2\cfrac{n\pi}{3}),求数列\{b_n\}的前n项和T_n$
$解:\cfrac{S_{n+1}}{S_n}=\cfrac{n+2}{n}\Rightarrow \cfrac{S_{n+1}}{S_n}-1=\cfrac{n+2}{n}-1\Rightarrow \cfrac{a_{n+1}}{S_n}=\cfrac{2}{n}\Rightarrow na_{n+1}=2S_n\quad {\color{Red} ①}$
$由递推公式得到(n-1)a_n=2S_{n-1}\quad {\color{Red} ②} \Rightarrow {\color{Red}①-② } ,得na_{n+1}-(n-1)a_n=2a_n$
$na_{n+1}=(n+1)a_n\Rightarrow \cfrac{a_{n+1}}{n+1}=\cfrac{a_{n}}{n}=\cfrac{a_1}{1}=1\Rightarrow a_n=n$
${\color{Red}方法2}:S_1\times \cfrac{S_2}{S_1}\times \cfrac{S_3}{S_2}\times \cfrac{S_4}{S_3}\times \cfrac{S_5}{S_4}\times\cdots\times \cfrac{S_{n+1}}{S_n}=1\times\cfrac{3}{1}\times\cfrac{4}{2} \times\cfrac{5}{3} \times\cfrac{6}{4} \times\cfrac{7}{5}\times \cdots\times \cfrac{n}{n-2}\times\cfrac{n+1}{n-1}\times\cfrac{n+2}{n} $
$S_{n+1}=\cfrac{(n+1)(n+2)}{2}, S_{n}=\cfrac{n(n+1)}{2},n\ge 2时，a_n=S_{n}-S_{n-1}=n,显然n=1也成立。$

$b_n=a_n^2(\cos^2\cfrac{n\pi}{3}-\sin^2\cfrac{n\pi}{3})=n^2\cos \cfrac{2n\pi}{3}$
$b_1=\cos \cfrac{2\pi}{3}=1^2\times(-\cfrac{1}{2})$
$b_2=2^2\times\cos \cfrac{4\pi}{3}=2^2\times(-\cfrac{1}{2})$
$b_3=3^3\times\cos \cfrac{6\pi}{3}=3^2\times(+1)$
$b_4=4^2\times\cos \cfrac{8\pi}{3}=4^2\times(-\cfrac{1}{2})$
$\cdots\cdots\cdots\cdots\cdots$
$①设n=3k时，T_n=-{\color{Red}\cfrac{1}{2}}\times\cfrac{3k(3k+1)(6k+1)}{6} +{\color{Red}\cfrac{3}{2}}\times 9\times \cfrac{k(k+1)(2k+1)}{6}=$
$=-\cfrac{1}{4}k(3k+1)(6k+1)+\cfrac{9}{4}k(k+1)(2k+1)=k[-\cfrac{1}{4}(3k+1)(6k+1)+\cfrac{9}{4}(k+1)(2k+1)]$
$=k[-\cfrac{18}{4}k^2-\cfrac{9}{4}k-\cfrac{1}{4}+\cfrac{18}{4}k^2+\cfrac{9}{4}\times 3k+\cfrac{9}{4}]$
$=k(\cfrac{9}{2}k+2)=\cfrac{9}{2}k^2+2k=\cfrac{9}{2}(\cfrac{n}{3})^2+\cfrac{2n}{3}=\cfrac{3n^2+4n}{6}$
$设②n=3k+1,T_n=\cfrac{9}{2}k^2+2k-{\color{Red}\cfrac{1}{2}(3k+1)^2}=\cfrac{9}{2}k^2+2k-\cfrac{1}{2}(9k^2+6k+1)$
$=-k-\cfrac{1}{2}\quad k=\cfrac{n-1}{3}\Rightarrow T_n=-\cfrac{n-1}{3}-\cfrac{1}{2}=-\cfrac{2n+1}{6}$
$③设n=3k+2,T_n=-k-\cfrac{1}{2}-\cfrac{1}{2}(3k+2)^2=-k-\cfrac{1}{2}-\cfrac{1}{2}\times n^2=-\cfrac{3n^2+2n+1}{6}$
${\color{Green}这里是错的,k值不同于 ②} $
$k=\cfrac{n-2}{3}\quad T_n=-k-\cfrac{1}{2}-\cfrac{1}{2}(3k+2)^2=-\cfrac{n-2}{3}-\cfrac{1}{2}-\cfrac{1}{2}\times n^2=-\cfrac{3n^2+2n-1}{6}$
$综合上述T_n=\begin{cases} \cfrac{3n^2+4n}{6}\qquad\quad n=3k时\\ -\cfrac{2n+1}{6}\qquad\quad n=3k+1时\\-\cfrac{3n^2+2n-1}{6}\quad n=3k+2\end{cases}$

${\color{Red}利用完全立方差或立方差公式推导a_n=n^2的前n项和S_n } $
$(n-1)^3＝n^2-3n^2+3n-1\Rightarrow n^3-(n-1)^3＝3n^2-3n+1$
$n^3-(n-1)^3＝3n^2-3n+1$
$(n-1)^3-(n-2)^3＝3(n-1)^2-3n(n-1)+1$
$(n-2)^3-(n-3)^3＝3(n-2)^2-3n(n-2)+1$
$(n-3)^3-(n-4)^3＝3(n-3)^2-3n(n-3)+1$
$\cdots \cdots \cdots$
$2^3-1^3＝3\cdot 2^2-3\times 2+1$
$1^3-0^3＝3\cdot 1^2-3\times 1+1$
左右两边相加
$n^3=3S_n-3\times \cfrac{n(n+1)}{2}+n$
$3S_n=n^3+3\times \cfrac{n(n+1)}{2}-n=n[n^2+\cfrac{3}{2}(n+1)-1]$
$\cfrac{n}{2}(2n^2+3n+1)=\cfrac{n}{2}(2n+1)(n+1)$
$S_n=\cfrac{1}{6}\times n(n+1)(2n+1)$

数列讨论奇偶求通项－构造相邻项

$21年新高考一卷，已知数列\{a_n\}满足a_1=1,a_{n+1}=\begin{cases}a_n+1,n为奇数\\a_n+2,n为偶数\end{cases}$
$(1)记b_n=a_{2n},写出b_1,b_2,并求数列\{b_n\}的通项公式；$
$(2)求\{a_n\}的前20项和$
$a_2=a_1+1=1+1=2,b_1=a_2=2,a_3=a_2+2=4,b_2=a_4=a_3+1=5$
${\color{Green} \because \quad b_n=a_{2n}}$
$\therefore b_{n+1}=a_ {2(n+1)}=a_ {{\color{Green} 2n+1} +1}=a_ {{\color{Red} 2n}+1}+1=a_{{\color{Red} 2n} }+2+1=b_n+3$
$\Rightarrow b_{n+1}=b_n+3,故b_n为首项为2,公差为3的等差数列,b_n=2+3(n-1)=3n-1$
$(2)求前20项和，需分别计算奇数项和偶数项和；$
$偶数项：共10项(a_2,a_4,a_6\cdots a_{20}),b_{10}=3\times 10-1=29,$
$S_偶=\cfrac{10(b_1+b_{10})}{2}=\cfrac{10(2+29)}{2}=155$
$奇数项：共10项(a_1,a_3,a_5\cdots a_{19})$
$设\{c_k\},(c_k=a_{2k-1},k\in 1,2,3,\cdots 10)$
$c_{k+1}=a_{2k+1}=a_{2k}+2=(a_{2k-1}+1)+2=a_{2k-1}+3=c_k+3$
$\{c_k\}是首项为c_1=1,公差d=3的等差数列，和为S_奇=\cfrac{10(C_1+C_{10})}{2}=145$

$例6.记S_n为正项数列\{a_n\}的前n项和，且a_1=\cfrac{1}{3},2S_n=(3^n-1)a_n;$
$(1)求数列\{a_n\}的通项公式;$
$(2)设b_n=\cfrac{a_{n+1}}{(1-a_n)S_{n+1}},记数列\{b_n\}的前n项和为T_n,证明:T_n\ge \cfrac{3}{8}$
$解:2S_n=(3^n-1)a_n\quad ① n\ge 2时，2S_{n-1}=(3^{n-1}-1)a_{n-1}\quad ②$
$①-②,得2a_n=(3^n-1)a_n-(3^{n-1}-1)a_{n-1}\Rightarrow (3^{n-1}-1)a_{n-1}=(3^n-3)a_n$
$\Rightarrow \cfrac{a_n}{a_{n-1}}=\cfrac{3^{n-1}-1}{3^n-3}=\cfrac{1}{3},所以a_n为首项\cfrac{1}{3}，公式q为\cfrac{1}{3}的等比数列$
$a_n=(\cfrac{1}{3})^n,S_n=\cfrac{\cfrac{1}{3}-(\cfrac{1}{3})^{n+1}}{1-\cfrac{1}{3}}=\cfrac{1-(\cfrac{1}{3})^n}{2}$
$b_n=\cfrac{a_{n+1}}{(1-a_n)S_{n+1}}=\cfrac{(\cfrac{1}{3})^{n+1}}{[1-(\cfrac{1}{3})^{n}]\cfrac{1-(\cfrac{1}{3})^{n+1}}{2}}=\cfrac{2\times (\cfrac{1}{3})^{n+1}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
${\color{Orange} \because \quad \cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}} =\cfrac{1-(\cfrac{1}{3})^{n+1}-[{1-(\cfrac{1}{3})^{n}}]}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
$=\cfrac{-(\cfrac{1}{3})^{n+1}{+(\cfrac{1}{3})^{n}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}=\cfrac{{\cfrac{2}{3}\times (\cfrac{1}{3})^{n}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}=\cfrac{{2\times (\cfrac{1}{3})^{n+1}}}{[1-(\cfrac{1}{3})^{n}][1-(\cfrac{1}{3})^{n+1}]}$
${\color{Orange} \therefore \quad b_n=\cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}} $
$T_n=b_1+b_2+b_3+\cdots +b_n={\color{Orange} \cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^2}}+\cfrac{1}{1-(\cfrac{1}{3})^{2}}-\cfrac{1}{1-(\cfrac{1}{3})^{3}}$
$+{\color{Orange} \cfrac{1}{1-(\cfrac{1}{3})^{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{4}}} +\cfrac{1}{1-(\cfrac{1}{3})^{4}}-\cfrac{1}{1-(\cfrac{1}{3})^{5}}+\cdots +\cfrac{1}{1-(\cfrac{1}{3})^{n}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}$
$=\cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}},要证\cfrac{3}{2}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}\ge \cfrac{3}{8}即证\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}\le \cfrac{9}{8}$
$即证1-(\cfrac{1}{3})^{n+1}\ge \cfrac{8}{9}=1-\cfrac{1}{9},n=1即可$
$T_n=\cfrac{1}{1-\cfrac{1}{3}}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}}=\cfrac{3}{2}-\cfrac{1}{1-(\cfrac{1}{3})^{n+1}},T_n\nearrow,T_n\ge (T_n)_{min}=T_1=\cfrac{3}{2}-\cfrac{9}{8}=\cfrac{3}{8}$

$例7、数列\{a_n\}的前n项和为S_n,且a_1=１,\{\cfrac{S_n}{a_n}\}是公差为\cfrac{1}{3}的等差数列$
$(1)求\{a_n\}通项公式$
$(2)证明:\cfrac{1}{a_1}+\cfrac{1}{a_2}+\cdots+\cfrac{1}{a_n}\lt 2$
$解:\cfrac{S_n}{a_n}=1+\cfrac{1}{3}(n-1)=\cfrac{n+2}{3}\Rightarrow S_n=\cfrac{n+2}{3}a_n\quad ①$
$n\ge 2时，有S_{n-1}=\cfrac{n-1+2}{3}a_{n-1}=\cfrac{n+1}{3}a_{n-1}\quad②$
$①-②,得\cfrac{a_n}{a_{n-1}}=\cfrac{n+1}{n-1}$
$\cfrac{a_2}{a_1}=\cfrac{2+1}{2-1}=\cfrac{3}{1}$
$\cfrac{a_3}{a_2}=\cfrac{3+1}{3-1}=\cfrac{4}{2}$
$\cfrac{a_4}{a_3}=\cfrac{4+1}{4-1}=\cfrac{5}{3}$
$\cdots \cdots\cdots\cdots $
$\cfrac{a_{n-1}}{a_{n-2}}=\cfrac{n-1+1}{n-1-1}=\cfrac{n}{n-2}$
$\cfrac{a_n}{a_{n-1}}=\cfrac{n+1}{n-1}$
$\cfrac{a_n}{a_{n-1}}\times\cfrac{a_{n-1}}{a_{n-2}}\cdots \cfrac{a_4}{a_3}\times \cfrac{a_3}{a_2}\times \cfrac{a_2}{a_1}=a_n=\cfrac{n+1}{n-1}\times\cfrac{n}{n-2}\cdots\cfrac{5}{3}\times \cfrac{4}{2}\times \cfrac{3}{1}=\cfrac{n(n+1)}{2}$
${\color{Red} \because \quad} \cfrac{1}{a_n}=\cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
${\color{Red} \therefore\quad} \cfrac{1}{a_1}+\cfrac{1}{a_2}+\cdots+\cfrac{1}{a_n}=2(1-\cfrac{1}{2}+\cfrac{1}{2}-\cfrac{1}{3}+\cdots +\cfrac{1}{n}-\cfrac{1}{n+1})=2(1-\cfrac{1}{n+1})\lt 2$

$例8、已知正项数列\{a_n\}的前n项和S_n，满足S_n=(\cfrac{a_n+1}{2})^2.$
$(1)求数列\{a_n\}的通项公式;$
$(2)记b_n=\cfrac{n+1}{S_nS_{n+2}},设数列\{b_n\}的前n项和为T_n.求证T_n\lt \cfrac{5}{16}$
$解：(1).a_1=S_1=(\cfrac{a_1+1}{2})^2\Rightarrow a_1=1$
$n\ge 2时，S_{n-1}=(\cfrac{a_{n-1}+1}{2})^2$
$a_n=S_n-S_{n-1}=(\cfrac{a_n+1}{2})^2-(\cfrac{a_{n-1}+1}{2})^2=\cfrac{a_n+a_{n-1}+2}{2} \times\cfrac{a_n-a_{n-1}+2}{2}$
$\Rightarrow 4a_n=a_n^2-a_{n-1}^2+2(a_n-a_{n-1})\Rightarrow a_n^2-a_{n-1}^2-2(a_n+a_{n-1})=0$
$\Rightarrow (a_n+a_{n-1})(a_n-a_{n-1}-2)=0,a_n+a_{n-1}=0(舍去),a_n-a_{n-1}=2$
$a_{n-1}-a_{n-2}=2$
$a_{n-2}-a_{n-3}=2$
$a_2-a_1=2$
$a_n-a_1=2(n-1)\Rightarrow a_n=2n-1\quad(n\ge 2),a=1$
$(2){\color{Red} \because \quad } S_n=(\cfrac{a_n+1}{2})^2=n^2,所以S_{n+2}=(n+2)^2$
$b_n=\cfrac{n+1}{S_nS_{n+2}}=\cfrac{n+1}{n^2(n+2)^2}$
${\color{Red} \because \quad } \cfrac{1}{n^2(n+2)^2} =\cfrac{1}{4(n+1)}[\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2} ]$
${\color{Red} \therefore \quad } b_n=\cfrac{n+1}{n^2(n+2)^2} =\cfrac{1}{4}[\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2} ]$
$T_n=\cfrac{1}{4}(1-\cfrac{1}{3^2}+\cfrac{1}{2^2}-\cfrac{1}{4^2}+\cfrac{1}{3^2}-\cfrac{1}{5^2}+\cdots+\cfrac{1}{n^2}-\cfrac{1}{(n+2)^2}$
$=\cfrac{1}{4}[1+\cfrac{1}{2^2}-\cfrac{1}{(n+1)^2}-\cfrac{1}{(n+2)^2}]\lt \cfrac{5}{16}$

https://uu.890222.xyz/usr/uploads/2025/12/705270186.[5]: https://uu.890222.xyz/usr/uploads/2025/12/2791294339.png

一、阶差法：

典例 $证明:\ln (n+1)+\cfrac{n}{2(n+1)}\lt 1+\cfrac{1}{2}+\cfrac{1}{3}+\cdots ++\cfrac{1}{n}$
$解答:设S_n=\ln (n+1)+\cfrac{n}{2(n+1)},a_n=\ln \cfrac{n+1}{n}+\cfrac{n}{2(n+1)}-\cfrac{n-1}{2n}$
$要证\ln (n+1)+\cfrac{n}{2(n+1)}\lt 1+\cfrac{1}{2}+\cfrac{1}{3}+\cdots ++\cfrac{1}{n},即证\ln \cfrac{n+1}{n}+\cfrac{n}{2(n+1)}-\cfrac{n-1}{2n}\lt \cfrac{1}{n}$
$整理得\ln \cfrac{n+1}{n}\lt \cfrac{1+n}{2n}-\cfrac{n}{2(n+1)}$
$令x=\cfrac{n+1}{n}\gt 1,构造\ln x\lt \cfrac{1}{2}(x-\cfrac{1}{x}),易证其恒成立。$
$在数列中我们学过求通项的方法：阶差法，即S_n-S_{n-1}=a_n.$
$在数列不等式的证明题中，我们可以将一串不等式为通项的前项和，从而在本题中可以$
$设S_n=\ln (n+1)+\cfrac{n}{2(n+1)},由阶差法法计算得a_n=\ln \cfrac{n+1}{n}+\cfrac{n}{2(n+1)}-\cfrac{n-1}{2n}，$
$同理可得前n项和1+\cfrac{1}{2}+\cfrac{1}{3}+\cdots +\cfrac{1}{n}的通项公式为\cfrac{1}{n},因此证明\ln \cfrac{n+1}{n}+\cfrac{n}{2(n+1)}-\cfrac{n-1}{2n}\lt \cfrac{1}{n}即可$
这种方法证明数列不等式是非常常用的一种。
$精练2：n\in N^*，证明:\cfrac{1}{\sqrt{1^2+1}}+\cfrac{1}{\sqrt{2^2+1}}+\cfrac{1}{\sqrt{3^2+1}}+\cfrac{1}{\sqrt{4^2+1}}+\cdots+\cfrac{1}{\sqrt{n^2+n}}\gt \ln (n+1)$
此题为2022年高考2卷22题第三问。
$原题:已知f(x)=xe^{ax}-e^x.$
$(1)当a=1时,f(x)的单调性;$
$(2)当x\gt 0时，f(x)\lt -1,求a的取值$
$(3)n\in N^*，证明:\cfrac{1}{\sqrt{1^2+1}}+\cfrac{1}{\sqrt{2^2+1}}+\cfrac{1}{\sqrt{3^2+1}}+\cfrac{1}{\sqrt{4^2+1}}+\cdots+\cfrac{1}{\sqrt{n^2+n}}\gt \ln (n+1)$
$此题还可以扩展为:$
$(1)当a=1时,f(x)的单调性;$
$(2)证明当x\gt 0时，总有\ln (x+1)\lt x;$
$(3)若0\lt a\le \cfrac{1}{2},求证:(1+ax)e^{ax}-e^x\le0$;
$(4)当x\gt 0时，f(x)\lt -1,求a的取值$
$(5)在(3)的条件下,当a=\cfrac{1}{2}时，xe^{\cfrac{1}{2}x}-e^x+1\lt 0,令t=e^{\cfrac{1}{2}x},化简上式并用对数式表示;$
$(6)求证:\ln (n+1)-\ln n\lt \cfrac{1}{\sqrt{n^2+n}}(n\in N^*);$
$(7)n\in N^*，证明:\cfrac{1}{\sqrt{1^2+1}}+\cfrac{1}{\sqrt{2^2+1}}+\cfrac{1}{\sqrt{3^2+1}}+\cfrac{1}{\sqrt{4^2+1}}+\cdots+\cfrac{1}{\sqrt{n^2+n}}\gt \ln (n+1)$

$设S_n=\ln (n+1),a_n=S_n-S_{n-1}=\ln (n+1)-\ln n即证\cfrac{1}{\sqrt{n^2+n}}\gt\ln (n+1)-\ln n $
利用对数均值不等式，易得证
$\cfrac{n+1-n}{\ln (n+1)-\ln n }\gt \sqrt{(n+1)n}\Rightarrow \cfrac{1}{\sqrt{n^2+n}}\gt\ln (n+1)-\ln n$

二、放缩法：
$放缩法常用式:e^x\ge x+1;x-1\ge \ln x;x\ge \sin x,x\in [0,+\infty)$
$典例(全国卷)已知函数f(x)=x-1-a\ln x.$
$(1)若f(x)\ge 0,求a的值.$
$(2)设m为整数，且对于任意正整数n,(1+\cfrac{1}{2})(1+\cfrac{1}{2^2})\cdots 1+\cfrac{1}{2^n}\lt m,求m的最小值.$
$解答:(1)f(x)的定义域为(0,+\infty).$
$①若a\le 0,因为f(\cfrac{1}{2} )=-\cfrac{1}{2} +a\ln x\lt 0,故不满足题意；$
$②若a\gt 0,由{f}' (x)=1-\cfrac{a}{x}知当x\in (0,a),{f}' (x)\lt 0;当x\in (a,+\infty)时，{f}' (x)\gt 0,$
$所以f(x)在(0，a)上\searrow;在(a,+\infty)上\nearrow.故当x=a是f(x)在(0,+\infty )上的最小值点。$
$由f(1)=0,因此，当且位仅当　a=1时f(x)\ge 0.故a=1.$
$(2)由(1)知当x\in(1,+\infty),x-1-\ln x\gt 0.$
$令x=1+\cfrac{1}{2^n},得\ln (1+\cfrac{1}{2^n)} \lt \cfrac{1}{2^n}$
$从而\ln (1+\cfrac{1}{2})+\ln (1+\cfrac{1}{2^2})+\cdots+\ln (1+\cfrac{1}{2^2})\lt \cfrac{1}{2}+\cfrac{1}{2^2}+\cdots +\cfrac{1}{2^n}=1-\cfrac{1}{2^n}\lt 1$
$故(1+\cfrac{1}{2})(1+\cfrac{1}{2^2})\cdots (1+\cfrac{1}{2^n})\lt e,而(1+\cfrac{1}{2})(1+\cfrac{1}{2^2})(1+\cfrac{1}{2^3})\gt 2$
$所以m的最小值为3$

$好题精练1:证明n\ge 2,n\in N^*,时(1+\cfrac{1}{2^2})(1+\cfrac{1}{3^2})(1+\cfrac{1}{4^2})\cdots (1+\cfrac{1}{n^2})\lt 1$
$分析:两边取自然对数,\ln (1+\cfrac{1}{2^2})+\ln (1+\cfrac{1}{3^2})+\ln (1+\cfrac{1}{4^2})+\cdots+\ln (1+\cfrac{1}{n^2})\lt 1$
$\ln (1+x)\le x恒成立，当且仅当x=1取=,\cfrac{1}{n^2}\lt \cfrac{1}{(n-1)n}$
$解:\ln (1+\cfrac{1}{2^2})+\ln (1+\cfrac{1}{3^2})+\ln (1+\cfrac{1}{4^2})+\cdots+\ln (1+\cfrac{1}{n^2})\lt\cfrac{1}{2^2}+\cfrac{1}{3^2}+\cfrac{1}{4^2}+\cdots +\cfrac{1}{n^2}$
$\lt \cfrac{1}{1\times 2}+\cfrac{1}{2\times 3}+\cfrac{1}{3\times 4}+\cdots +\cfrac{1}{(n-1)n}=1-\cfrac{1}{n}\lt 1$

$好题精练2:已知函数f(x)=(x+2)\ln (x+2),g(x)=x^2+(3-a)x+2(1-a)(a\in R)$
$(1)函数f(x)的极值。$
$(2)若不等式f(x)\ge g(x)在(-2,+\infty)上恒成立,求a的取值范围$
$(3)证明:(1+\cfrac{1}{4})(1+\cfrac{1}{4^2})\cdots (1+\cfrac{1}{4^n})\lt e^\frac{1}{3}(n\in N^*).$
$解:(1){f}' (x)=\ln (x+2)+1,由{f}' (x)\gt 0可得x\gt \cfrac{1}{e}-2,则f(x)在(\cfrac{1}{e}-2,+\infty)上单调递增；$
$由{f}' (x)\lt 0可得x\lt \cfrac{1}{e}-2,则f(x)在(-2,\cfrac{1}{e}-2)上单调递减。$
$所以当x= \cfrac{1}{e}-2时,f(x)取得极小值，极小值为-\cfrac{1}{e},无极大值.$
$(2)分离参数后，构造函数，通过求函数的最小值确定参数a的取值范围.$
$由不等式f(x)\ge g(x)在(-2,+\infty)上恒成立,得(x+2)\ln (x+2)\ge x^2+(3-a)x+2(1-a)$
$(x+2)\ln (x+2)\ge (x+2)(x+1-a),x\in (-2,+\infty),x+2\gt 0$
$a\le x+1-\ln(x+2)在x\in (-2,+\infty)上恒成立$
$设h(x)=x+1-\ln(x+2),{h}' (x)=1-\cfrac{1}{x+2}=\cfrac{x+1}{x+2},令{h}' (x)=0,得x=-1,$
$当x\in (-2,-1)时，{h}' (x)\lt 0,所以h(x)在(-2，-1)上单调递减；$
$当x\in (-1，+\infty)时，{h}' (x)\gt 0,所以h(x)在(-1，+\infty)上单调递增；$
$所以h(x)_{min}=h(-1)=0,则a\le 0,所以a的取值范围为(-\infty,0].$
$(3)令x+1=4^n,则\ln(1+\cfrac{1}{4^n})\lt \cfrac{1}{4^n}$
$所以\ln(1+\cfrac{1}{4})+\ln(1+\cfrac{1}{4^2})+\cdots +\ln(1+\cfrac{1}{4^n})\lt \cfrac{1}{4}+\cfrac{1}{4^2}+\cdots+\cfrac{1}{4^n}=\cfrac{1}{3}(1-\cfrac{1}{4^n}),$
$即\ln[(1+\cfrac{1}{4})(1+\cfrac{1}{4^2})\cdots(1+\cfrac{1}{4^n})]\lt \cfrac{1}{3}(1-\cfrac{1}{4^n})$

三、裂项法：

$\qquad\lt \cfrac{2}{e}(1^2+\cfrac{1}{2^2}+\cfrac{1}{3^2} +\dots+\cfrac{1}{n^2} ) $

四、构造函数：

五、最值型证明

六、累加型证明
-------

七、数学归纳法：

1、裂项处理：
例1、2018年天津理科数学18题（2）
$证明：\sum_{n}^{k=1}\cfrac{k\cdot 2^{k+1}}{(k+1)(k+2)} =\cfrac{2^{n+2}}{n+2}-2$
设裂项的形式为: $\cfrac{a\cdot 2^{k+2}}{k+2}- \cfrac{a\cdot2^{k+1}}{k+1}=\cfrac{k\cdot 2^{k+1}}{(k+1)(k+2)}$
$代入得a=1,则a=1对应的表达式即为所求。$
分析：其实这题只是求和，但是很好的说明了一种探究数列裂项方式——待定系数.在解题时根据待证表达式的特征确定裂项的形式和需要待定的系数，再利用等号成立的条件对应系数得到结果. 对于涉及裂项的数列不等式问题，给出的数列求和往往极限是所需证明的结果.

$1、证明:{\color{Green} \cfrac{1}{1^2}+\cfrac{1}{2^2} +\cfrac{1}{3^2}+\dots ++\cfrac{1}{n^2}\lt 2} $
$2、证明:{\color{Green} \cfrac{1}{1^2}+\cfrac{1}{2^2} +\cfrac{1}{3^2}+\dots ++\cfrac{1}{n^2}\lt \cfrac{7}{4} } $
$3、证明:{\color{Green} \cfrac{1}{1^2}+\cfrac{1}{2^2} +\cfrac{1}{3^2}+\dots ++\cfrac{1}{n^2} \lt \cfrac{5}{3} }$
这数列显然是没办法直接求和的，欧拉已证明当$n\to +\infty时，级数\sum_{n}^{i=1} \cfrac{1}{i^2}=\cfrac{\pi^2}{6}$
$1、{\color{Green}\cfrac{1}{n^2}=\cfrac{1}{n\cdot n}\lt \cfrac{1}{n^2-n}=\cfrac{1}{n\cdot (n-1)}= } \cfrac{1}{n-1}- \cfrac{1}{n}(n\ge 2)$
$原式\lt 1+(1-\cfrac{1}{2})+(\cfrac{1}{2}-\cfrac{1}{3} )+(\cfrac{1}{3}-\cfrac{1}{4} )+\dots +(\cfrac{1}{n-1}-\cfrac{1}{n} )\lt 2-\cfrac{1}{n}\lt 2$
$2、{\color{Green}\cfrac{1}{n^2}\lt \cfrac{1}{n^2-1}=\cfrac{1}{2}(\cfrac{1}{n-1}-\cfrac{1}{n+1} )\quad (n\ge 2)}$
$原式\lt 1+\cfrac{1}{2}[(1-\cfrac{1}{3})+(\cfrac{1}{2}-\cfrac{1}{4})+ (\cfrac{1}{3}-\cfrac{1}{5}) +(\cfrac{1}{4}-\cfrac{1}{6})+\dots +(\cfrac{1}{n-2}-\cfrac{1}{n})-(\cfrac{1}{n-1}-\cfrac{1}{n+1})]$
$=1+\cfrac{1}{2}(1+\cfrac{1}{2}-\cfrac{1}{n}--\cfrac{1}{n+1}) \lt 1+\cfrac{3}{4}=\cfrac{7}{4} $
$3、 {\color{Green}\cfrac{1}{n^2}\lt \cfrac{1}{n^2-\cfrac{1}{4} }=\cfrac{1}{n-\cfrac{1}{2} }-\cfrac{1}{n+\cfrac{1}{2} } \quad n\ge 2}$
$原式=\color{Green} \cfrac{1}{1^2}+\cfrac{1}{2^2} +\cfrac{1}{3^2}+\dots +\cfrac{1}{n^2}\lt 1+(\cfrac{1}{2-\cfrac{1}{2}}-\cfrac{1}{2+\cfrac{1}{2}} )+(\cfrac{1}{3-\cfrac{1}{2}}-\cfrac{1}{3+\cfrac{1}{2}})(\cfrac{1}{4-\cfrac{1}{2}}-\cfrac{1}{4+\cfrac{1}{2}} )+\dots +(\cfrac{1}{n-\cfrac{1}{2}}-\cfrac{1}{n+\cfrac{1}{2}} )$
$= 1+(\cfrac{1}{\cfrac{3}{2} } -\cfrac{1}{\cfrac{5}{2} } )+(\cfrac{1}{\cfrac{5}{2} } -\cfrac{1}{\cfrac{7}{2} } )+(\cfrac{1}{\cfrac{7}{2} } -\cfrac{1}{\cfrac{9}{2} } )+\dots +(\cfrac{1}{\cfrac{2n-1}{2} } -\cfrac{1}{\cfrac{2n+1}{2} } )$
$= 1+(\cfrac{1}{\cfrac{3}{2} } -\cfrac{1}{\cfrac{2n+1}{2} } )\lt \cfrac{5}{3} $

$4、求证：\cfrac{1}{3} +\cfrac{1}{5} +\cfrac{1}{7} +\dots +\cfrac{1}{2n+1} \lt \ln (n+1)$
分析：按照常规做法，观察到左边共有$n项，于是我们尝试将右边的\ln(n+1)分解为一个n项式，再逐一比较。$
按常规拆分有：
$\ln (n+1)=\ln (\cfrac{n+1}{n})+\ln (\cfrac{n}{n-1})+\ln (\cfrac{n-1}{n-2})+\dots+\ln \cfrac{2}{1}$
只需要证明：$\ln (\cfrac{n+1}{n})\gt \cfrac{1}{2n+1}$
$\Leftrightarrow \ln (1+\cfrac{1}{n})\gt \cfrac{\cfrac{1}{n}}{2+\cfrac{1}{n}} $
$换元令x=\cfrac{1}{n}+1\quad x\gt 1,即证明:\ln x\gt \cfrac{x-1}{x+1}$
这里略去求导过程，直接用常规放缩即证得。
$\ln x\gt 2\cdot \cfrac{x-1}{x+1} \gt \cfrac{x-1}{x+1} \quad x\gt 1$

实际上，大多数列不等式的题都是像这样，将一项拆为多项再逐一比较大小。
回看反思这种做数列不等式题的常规方法，我想：我们很擅长比较两个初等函数函数的大小，求导即可。
把一项拆分成多项的原因是无法找到LHS的初等函数表达式。（何况LHS是发散的）

于是我们可以通过积分放缩来找到左边的近似函数。

积分放缩：

以此题为例， LHS每一项的数值都在函数$f(x)=\cfrac{1}{2n+1}$上。则：
$LHS=1\times f(1)+1\times f(2)+1\times f(3)+\dots +1\times f(n)$
$LHS\lt \int_{0}^{n} f(x)\mathrm{d}x=[\cfrac{1}{2}\ln (2n+1) ]_{0}^{n} = \cfrac{1}{2}\ln (2n+1) $
即证明：
$\cfrac{1}{2}\ln (2n+1)\lt \ln (n+1)$
$\ln (2n+1)\lt \ln (n+1)^2=\ln (n^2+2n+1)$
解答题当然不能这样写，写出来可能就得一分辛苦分。
但是部分小题可以用，比如下面这道题：

针对$a_{n+1}=f(a_n)=\cfrac{pa_b+q}{ra_n+s}\quad (p,s,r\ne 0)\quad $
${\color{Red}(1) } 若q=0,a_{n+1}=\cfrac{pa_b}{ra_n+s},\quad $倒数法
${\color{Red} (2) 不动点法}$
若$y=f(x)有f(x_0)=x_0,则x_0是y=f(x)$的不动点。
即法则$f仅对x_0$失效了，经过法则后，仍旧等于它本身。
${\color{Red} 不动点的性质：} f(x)-x=(x-x_0)\cdot A\quad$ (A是多项式)
证明：$\because x_0是y=f(x)的$不动点，
$\Rightarrow x_0是f(x)-x＝0$的根。
$\Rightarrow f(x)-x=(x-x_0)\cdot A$
针对数列$a_{n+1}=f(a_n),\quad若x_0是y=f(a_n)$的不动点，
$a_{n+1}-x_0=f(a_n)-x_0=(a_n-x_0)\cdot A$
$a_{n+1}=f(a_n)=\cfrac{pa_b+q}{ra_n+s}$对应特征方程：$x=\cfrac{px+q}{rx+s}$
若对应特征方程：$x=\cfrac{px+q}{rx+s}$有以下情况：
${\color{Red} ①} 有两个不相等的实根\alpha和 \beta ,则\{\cfrac{a_n-\alpha }{a_n-\beta } \}$为等比数列；
${\color{Red} ②} 有两个相等的实根\alpha，则\{\cfrac{1}{a_n-\alpha} \}$为等差数列；
${\color{Red} ③}$ 没有实根，则原$a_n$是周期数列。