6. Given the general term of the sequence $\left\{a_{n}\right\}$
$$
a_{n}=\frac{n x}{(x+1)(2 x+1) \cdots(n x+1)}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

If $a_{1}+a_{2}+\cdots+a_{2015}<1$, then the value of the real number $x$ is $(\quad)$.
(A) $-\frac{3}{2}$
(B) $-\frac{5}{12}$
(C) $-\frac{9}{40}$
(D) $-\frac{11}{60}$
6. D.

From the problem, we know
$$
\begin{array}{l}
a_{n}=\frac{1}{(x+1)(2 x+1) \cdots[(n-1) x+1]}- \\
\frac{1}{(x+1)(2 x+1) \cdots(n x+1)} . \\
=1-\frac{1}{(x+1)(2 x+1) \cdots(2015 x+1)}0 . \\
\sum_{k=1}^{2015} a_{k} \\
\text { Therefore, } x \in \bigcup_{k=1}^{100}\left(-\frac{1}{2 k-1},-\frac{1}{2 k}\right) \cup\left(-\frac{1}{2015},+\infty\right) .
\end{array}
$$

Upon verification, only $x=-\frac{11}{60}$ satisfies the problem.