Question

Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent?

\(\sum\limits_1^\infty {\frac{{{{( - 1)}^{n - 1}}}}{{n!}}} \)

Short answer

The first eights terms of partial sum are \(1,\frac{{ - 1}}{2},\frac{1}{6},\frac{{ - 1}}{{24}},\frac{1}{{120}},\frac{{ - 1}}{{720}},\frac{1}{{5760}},\frac{{ - 1}}{{51840}}\)…And the series is convergent.

Step-by-step solution

Using Leibnitz test ,Condition-1

\({u_n} > {u_{n + 1}}\)

\({u_n} = \frac{{{{( - 1)}^{n - 1}}}}{{n!}}\),\({u_n} > \frac{1}{{n!}}\)

\({u_n} > {u_{n + 1}}\)

Finding the limit

\({\lim _{n \to \infty }}{u_n} = {\lim _{n \to \infty }}\frac{1}{{n!}} = 0\)

Hence, The series is convergent by Leibnitz Test.