題目：
設數列 \(\left<a_n\right>\) 的前 \(n\) 項和為 \(S_n\) \((n\geq 2)\)
\(a_1=1\)，若 \(3S_n^2 = a_n\left(3S_n-1\right)\)
(1) 試證：數列 \(\displaystyle \left<\frac{1}{S_n}\right>\) 為一等差數列
(2)若 \(\displaystyle b_n=\frac{S_n}{3n+1}\) 且 \(b_n\) 的前 \(n\) 項和為 \(T_n\)，求 \(\displaystyle \lim_{n\to\infty} T_n=?\).
解答：
(1)
定義 \(S_1=a_1\)
當 \(n\geq2\) 時，
\(3S_n^2 = a_n\left(3S_n-1\right)\)
\(\displaystyle \Rightarrow 3S_n^2 = \left(S_n-S_{n-1}\right)\left(3S_n-1\right)\)
\(\displaystyle \Rightarrow 3S_n^2 = 3S_n^2-3S_nS_{n-1}+S_{n-1}-S_n\)
\(\displaystyle \Rightarrow S_{n-1}-S_n=3S_nS_{n-1}\)
\(\displaystyle \Rightarrow \frac{1}{S_n}-\frac{1}{S_{n-1}}=3,\forall\, n\geq 2\)
\(\displaystyle \left<\frac{1}{S_n}\right>\) 是首項為 \(1\)，公差為 \(3\) 的等差數列。
(2)
當 \(n\) 為正整數時，
\(\displaystyle \frac{1}{S_n}=\frac{1}{S_1}+\left(n-1\right)\cdot3=1+\left(n-1\right)\cdot3=3n-2\)
\(\displaystyle b_n=\frac{S_n}{3n+1}=\frac{1}{\left(3n+1\right)\left(3n-2\right)}=\frac{1}{3}\left(\frac{1}{3n-2}-\frac{1}{3n+1}\right)\)
\(\displaystyle T_n=\sum_{k=1}^n b_k=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{3n+1}\right)\)
\(\displaystyle \Rightarrow \lim_{n\to\infty} T_n=\frac{1}{3}\)
