数列

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

$(1)a_{n+1}=a_n+n+1，有递推关系式：$
$a_{n}=a_{n-1}+(n-1)+1$
$a_{n-1}=a_{n-2}+(n-2)+1$
$a_{n-2}=a_{n-3}+(n-3)+1$
$a_{n-3}=a_{n-4}+(n-4)+1$
$\cdots \cdots \cdots\cdots$
$a_3=a_2+2+1$
$a_2=a_1+1+1$
累加之
$a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\cdots +a_3+a_2=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4}+\cdots +a_2+a_1$
$+(n-1)+(n-2)+(n-3)+(n-4)+\cdots +1+1+1\cdots +1$
$a_n=a_1+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1(n-1)\times 1$
$=2+(n-1)+(n-2)+(n-3)+(n-4)+\cdots+2+1+(n-1)$
$=n+(n-1)+\cdots +1=\cfrac{n(n+1)}{2}$

$b_n=\cfrac{1}{a_n}=\cfrac{1}{\cfrac{n(n+1)}{2}}=\cfrac{2}{n(n+1)}=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$b_1=2(1-\cfrac{1}{2})$
$b_2=2(\cfrac{1}{2}-\cfrac{1}{3})$
$b_3=2(\cfrac{1}{3}-\cfrac{1}{4})$
$\cdots\cdots\cdots\cdots$
$b_{n-2}=2(\cfrac{1}{n-2}-\cfrac{1}{n-1})$
$b_{n-1}=2(\cfrac{1}{n-1}-\cfrac{1}{n})$
$b_n=2(\cfrac{1}{n}-\cfrac{1}{n+1})$
$T_n=2(1-\cfrac{1}{n+1})\lt 2$

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

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