$识别\Rightarrow 裂项\Rightarrow 转化\Rightarrow 放缩\Rightarrow 求和$
$1.已知函数f(x)=\ln (x+1),g(x)=ax^2+x.$
$(1)求函数f(x)的图像在点(1，f(1))处的切线方程;$
$(2).当x\gt -1时，f(x)\le g(x)，求实数a的取值范围;$
$(3).已知n\in N^*,证明:\sin \cfrac{1}{n+1}+\sin \cfrac{1}{n+2}+\cdots +\sin \cfrac{1}{2n}\ln 2$
$1.识题型，定目标,左边累加，右边对数$
$2.对数裂项,\ln 2=\ln (\cfrac{n+1}{n}\times \cfrac{n+2}{n+1}\times \cfrac{n+3}{n+2}\times\cdots\times\cfrac{2n}{2n-1} )$
$3.证明左边每一项与右边每一项比大小$
$4.单项放缩:\sin \lt x,\ln x\gt 1-\cfrac{1}{x},\ln x\le x-1$
$5.求和$
$解: \ln 2=\ln (\cfrac{n+1}{n}\times \cfrac{n+2}{n+1}\times \cfrac{n+3}{n+2}\times\cdots\times\cfrac{2n}{2n-1}=\ln \cfrac{n+1}{n}+\ln \cfrac{n+2}{n+1}+\ln \cfrac{n+3}{n+2}+\cdots \ln \cfrac{2n}{2n-1}$
$$\sum_{k=n}^{2n-1}\sin \cfrac{1}{k+1}\lt \sum_{k=n}^{2n-1}\ln \cfrac{k+1}{k}$$
$\sin \cfrac{1}{k+1}\lt \cfrac{1}{k+1}=\cfrac{k+1-k}{k+1}=1-\cfrac{k}{k+1}\lt \ln \cfrac{k+1}{k}$

$2.已知函数f(x)=\ln x-ax^2+(2-a)x.$
$(1)若函数f(x)在[1,+\infty)上为减函数,求a的取值范围;$
$(2)当a=1时,g(x)＝x^2-2x+b,当x\in[\cfrac{1}{2},2)时,f(x)与g(x)有两个交点,求实数b的取值范围;$
$(3)证明:\cfrac{2}{1^2}+\cfrac{3}{2^2}+\cfrac{4}{3^2}+\cfrac{5}{4^2}+\cdots+\cfrac{n+1}{n^2}\gt \ln(n+1)\quad(\forall n\in N*).$
$上题有\ln 2\Rightarrow \ln \cfrac{2}{1}\Rightarrow\ln \cfrac{2n}{n}=\ln (\cfrac{\quad }{n}\qquad\qquad \cfrac{2n }{\quad})$
$本题:\ln (n+1)\Rightarrow \ln \cfrac{n+1}{1}=\ln (\cfrac{\quad }{1}\qquad\qquad \cfrac{n+1 }{\quad})$
$即证\cfrac{k+1}{k^2}\gt \ln \cfrac{k+1}{k},$
$令\cfrac{k+1}{k}＝x,\ln x\le x-1\Rightarrow \ln \cfrac{k+1}{k}\le \cfrac{k+1}{k}-1=\cfrac{1}{k}=\cfrac{k}{k^2}\lt \cfrac{k+1}{k^2}$