每日一题_191219

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.