14. For any $1 \leqslant k \leqslant n$, since $1+a_{1} \geqslant 2 \sqrt{a_{1}}, 1+a_{2} \geqslant 2 \sqrt{a_{2}}, \cdots, 1+$ $a_{k} \geqslant 2 \sqrt{a_{k}}$, noting that $a_{1} a_{2} \cdots a_{k} \geqslant 1$ we get
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{k}\right) \geqslant 2^{k} \sqrt{a_{1} a_{2} \cdots a_{k}} \geqslant 2^{k}$$

Thus,
$$\frac{1}{1+a_{1}}+\frac{2}{\left(1+a_{1}\right)\left(1+a_{2}\right)}+\cdots+\frac{n}{\left(1+a_{1}\right)\left(1+a_{2}\right)\left(1+a_{n}\right)} \leqslant \sum_{k=1}^{n} \frac{k}{2^{k}}$$

Let $S=\sum_{k=1}^{n} \frac{k}{2^{k}}$. Then
$$2 S=\sum_{k=1}^{n} \frac{k}{2^{k-1}}=\sum_{k=0}^{n-1} \frac{k+1}{2^{k}}$$

Therefore,
$$S=2 S-S=1-\frac{n}{2^{n}}+\sum_{k=0}^{n-1} \frac{1}{2 k}=2-\frac{n}{2^{n}}-\frac{1}{2^{n-1}}<2$$

Thus, we obtain
$$\sum_{k=1}^{n} \frac{k}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{k}\right)} \leqslant \sum_{k=1}^{n} \frac{k}{2^{k}}<2$$