Leibniz Gottfried Wilhelm

Consider a numerical triangle:

$$
\begin{array}{cccccc}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \ldots & \frac{1}{1993} \\
\frac{1}{2} & \frac{1}{6} & \frac{1}{12} & \ldots & \frac{1}{1992 \cdot 1993} \\
\frac{1}{3} & \frac{1}{12} & \ldots &
\end{array}
$$

(The first row is given, and each element of the subsequent rows is calculated as the difference of the two elements above it). In the 1993rd row, there is one element. Find it.
On the $j$-th place in the $i$-th row, the number is $\frac{(i-1)!(j-1)!}{(i+j-1)!}$ (this can be easily proven by induction on $\left.i\right)$. Substituting $j$ $=1, i=1993$, we get the desired number.

## Answer

$1 / 1993$.