11. Let the sequence $\left\{a_{n}\right\}$ have the sum of the first $n$ terms $S_{n}$ related to $a_{n}$ by $S_{n}=-b a_{n}+1-\frac{1}{(1+b)^{n}}$, where $b$ is a constant independent of $n$, and $b \neq-1$.
(1) Find the expression for $a_{n}$ in terms of $b$ and $n$;
(2) When $b=3$, from which term does the sequence $\left\{a_{n}\right\}$ satisfy $S_{n}-a_{n}>\frac{1}{2}$?
11. Analysis (1) $S_{n+1}=-b a_{n+1}+1-\frac{1}{(1+b)^{n+1}}, \quad S_{n}=-b a_{n}+1-\frac{1}{(1+b)^{n}}$ thus $a_{n+1}=-b a_{n+1}+b a_{n}+\frac{b}{(1+b)^{n+1}}$, which means $a_{n+1}=\frac{b}{1+b} a_{n}+\frac{b}{(1+b)^{n+2}}$; let $c_{n}=a_{n}(1+b)^{n+1}$, we get: $c_{n+1}=b c_{n}+b$ If $b=1$, then $a_{n}=\frac{n}{2^{n+1}}$; If $b \neq 1$, then $a_{n}=\frac{b\left(1-b^{n}\right)}{\left(1-b^{2}\right)(1+b)^{n}}$.
(2) When $b=3$, then $a_{n}=\frac{3\left(3^{n}-1\right)}{8 \times 4^{n}}$;

Since $S_{n}-a_{n}=-4 a_{n}+1-\frac{1}{4^{n}}=\frac{1-3^{n+1}}{2 \times 4^{n}}+1$, it is sufficient to have $4^{n}+1>3^{n+1} \quad(*)$.
Also, when $n \geq 4$, $4^{n}=(1+3)^{n} \geq 2+2 n 3^{n-1}+\frac{n(n-1)}{2} 3^{n-2}=2+\frac{n^{2}+11 n}{2} 3^{n-2}>3^{n+1}$, so we only need to consider the cases $n=1,2,3$. Upon inspection, $n=1,2,3$ do not satisfy (*), hence from the 4th term onwards, $S_{n}-a_{n}>\frac{1}{2}$.