6. Let $\left\{a_{n}\right\}$ be a sequence with the sum of the first $n$ terms denoted as $S_{n}$, and for any positive integer $n, S_{n}+a_{n}=1$. Let $S_{n}^{(1)}=S_{n}$, $S_{n}^{(k)}=S_{1}^{(k-1)}+S_{2}^{(k-1)}+\cdots+S_{n}^{(k-1)}(k \geqslant 2)$. Then $\lim _{n \rightarrow \infty} \frac{S_{n}^{(2015)}}{n^{2014}}=$ $\qquad$ .
6. $\frac{1}{2014!}$.

From the problem, we know $2 S_{1}=1$, i.e., $S_{1}=1, S_{n}+\left(S_{n}-S_{n-1}\right)=1 \Rightarrow S_{n}=\frac{1}{2}\left(S_{n-1}+1\right)$. It is easy to solve that $S_{n}=1-\frac{1}{2^{n}}$. Therefore,
$$
\begin{array}{l}
S_{n}^{(2)}=n-\left(1-\frac{1}{2^{n}}\right)=n-1+a_{n}, \\
S_{n}^{(3)}=\frac{1}{2}(n-1) n+S_{n}^{(1)}, \\
S_{n}^{(4)}=\frac{1}{6}(n-1) n(n+1)+S_{n}^{(2)}, \\
\cdots, \\
S_{n}^{(2015)}=\frac{1}{2014!}(n-1) n(n+1) \cdots(n+2012)+S_{n}^{(2013)} .
\end{array}
$$

Therefore, the required limit value is $\frac{1}{2014!}$.