Math521_刘雷

若数列$\{a_n\}$满足$a_{n+1}=3a_n+2$,$n\in\mathbb{N}^\ast$,且$a_1=2$.
$(1)$ 求数列$\{a_n\}$的通项公式;
$(2)$ 记$b_n=\dfrac{1}{a_n^2}+\dfrac{1}{a_n}$,数列$\{b_n\}$的前$n$项和为$S_n$,求证:$\forall n\in\mathbb{N}^\ast,S_n<\dfrac{31}{32}$.

解析:
$(1)$ 由题 $$a_{n+1}+1=3\left(a_n+1\right).$$ 所以$$
a_n=3^{n-1,n\in\mathbb{N}}\ast.$$
$(2)$ 结合$(1)$可知$b_n=\dfrac{3^n}{\left(3^n-1\right)^2}$,所以$$
\forall n\geqslant 2,b_n=\dfrac{3^{{n-1}}{\left(3}n-1\right)\left(3^{{n-1}-\dfrac{1}{3}\right)}<\dfrac{1}{2}\left(\dfrac{1}{3}-1}-\dfrac{1}{3^n-1}\right). $$从而$$
\begin{split}
S_n&\leqslant b_1+b_2+\sum_{k=3}^{n}\left[\dfrac{1}{2}\left( \dfrac{1}{3^{{k-1}-1}-\dfrac{1}{3}k-1}\right)\right]\
&=\dfrac{3}{4}+\dfrac{9}{64}+\dfrac{1}{16}-\dfrac{1}{2}\cdot\dfrac{1}{3^n-1}\
&<\dfrac{61}{64}<\dfrac{31}{32}.
\end{split}
$$
证毕.