4. $\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+3+\cdots+2016}\right)=$ $\qquad$

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4. $\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+3+\cdots+2016}\right)=$ $\qquad$
4. $\frac{1009}{3024}$.

From $1-\frac{1}{1+2+3+\cdots+n}=\frac{(n-1)(n+2)}{n(n+1)}$, we get
$$
\begin{aligned}
& \left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+3+\cdots+2016}\right) \\
= & \frac{1 \times 4}{2 \times 3} \cdot \frac{2 \times 5}{3 \times 4} \cdot \frac{3 \times 6}{4 \times 5} \cdots \cdots \frac{2015 \times 2018}{2016 \times 2017} \\
= & \frac{2018}{3 \times 2016}=\frac{1009}{3024} .
\end{aligned}
$$