#### Answer

The first four terms of the sequence with nth term ${{a}_{n}}=\frac{1}{\left( n-1 \right)!}$are$\underline{1,1,\frac{1}{2}\,\text{and}\,\frac{1}{6}}$.

#### Work Step by Step

In order for computing the \[1st\] four terms of the series having the general term as ${{a}_{n}}=\frac{1}{\left( n-1 \right)!}$, there is a need of replacing the n in the formula with \[1,2,3\,\text{and}\,4\].
1st term ${{a}_{1}}=\frac{1}{\left( 1-1 \right)!}=\frac{1}{0!}=\frac{1}{1}=1$
2nd term ${{a}_{2}}=\frac{1}{\left( 2-1 \right)!}=\frac{1}{1!}=\frac{1}{1!}=\frac{1}{1}=1$
3rd term ${{a}_{3}}=\frac{1}{\left( 3-1 \right)!}=\frac{1}{2!}=\frac{1}{2\times 1}=\frac{1}{2}$
4th term \[{{a}_{4}}=\frac{1}{\left( 4-1 \right)!}=\frac{1}{3!}=\frac{1}{3\times 2\times 1}=\frac{1}{6}\]
Hence, the first four terms of the sequence with nth term ${{a}_{n}}=\frac{1}{\left( n-1 \right)!}$are\[\underline{1,1,\frac{1}{2}\,\text{and}\,\frac{1}{6}}\].