4. Given the sequence $a_{1}=1, a_{n}=\frac{n}{n^{2}-1}(n \geqslant 2)$, then $\sum_{k=1}^{n}\left(a_{1} a_{2} \cdots a_{k}\right)=$
$$
a_{1} a_{2} \cdots a_{k}=\frac{2 \cdot 3 \cdots k}{1 \cdot 2 \cdot[3 \cdots(k-1)]^{2} \cdot k(k+1)}=\frac{2 k}{(k+1)!}=2\left[\frac{1}{k!}-\frac{1}{(k+1)!}\right] \text {, }
$$

So $\sum_{k=1}^{n}\left(a_{1} a_{2} \cdots a_{k}\right)=2\left[1-\frac{1}{(n+1)!}\right], n=1$ also satisfies the above equation.