Intermediate Algebra for College Students (7th Edition)

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