11. Given a sequence $\left\{a_{n}\right\}$ with all terms non-zero, the sum of the first $k$ terms is $S_{k}$, and $S_{k}=\frac{1}{2} a_{k} a_{k+1}\left(k \in \mathbf{N}_{+}\right)$, where $a_{1}=1$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) For any given positive integer $n(n \geqslant 2)$, the sequence $\left\{b_{n}\right\}$ satisfies: $\frac{b_{k+1}}{b_{k}}=\frac{k-n}{a_{k+1}}(k=1,2,3, \cdots, n), b_{1}=1$. Find $b_{1}+b_{2}+\cdots+b_{n}$.
(1) $\left\{\begin{array}{l}S_{k+1}=\frac{1}{2} a_{k+1} a_{k+2}, \\ S_{k}=\frac{1}{2} a_{k} a_{k+1}\end{array} \Rightarrow a_{k+1}=\frac{1}{2} a_{k+1}\left(a_{k+2}-a_{k}\right) \Rightarrow a_{k+2}-a_{k}=2\right.$, and $a_{k}=\frac{1}{2} a_{1} a_{2} \Rightarrow a_{2}=2$,

so $a_{n}=n\left(n \in \mathbf{N}_{+}\right)$.
$$
\begin{array}{l}
\text { (2) } \frac{b_{k}}{b_{k-1}}=-\frac{n-k+1}{k} \Rightarrow b_{k}=\left(\prod_{i=2}^{k} \frac{b_{i}}{b_{i-1}}\right) \cdot b_{1} \\
=(-1)^{k-1} \cdot \frac{(n-1)(n-2) \cdots(n-k+1)}{2 \cdot 3 \cdots k}=(-1)^{k-1} \frac{1}{n} C_{n}^{k} .
\end{array}
$$

So $b_{1}+b_{2}+\cdots+b_{n}=-\frac{1}{n}\left[-C_{n}^{1}+C_{n}^{2}-\cdots+(-1)^{k} C_{n}^{k}+\cdots+(-1)^{n} C_{n}^{n}\right]$
$$
=\frac{1}{n}-\frac{1}{n}\left[C_{n}^{0}-C_{n}^{1}+\cdots+(-1)^{k} C_{n}^{k}+\cdots+(-1)^{n} C_{n}^{n}\right]=\frac{1}{n} .
$$