In Exercises 21–28 provide the first five terms of the series.

$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$

$\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$

The first term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is obtained by substituting $n=0$in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=0$is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^0}{(2\times 0)!}$(Substituting)

=1 (Because 0!=1)

The first term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is 1 .

The second term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is obtained by substituting $n=1$in $\frac{(-1{)}^n}{\left(2n\right)!}$Therefore, the value at $n=1$is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^1}{(2\times 1)!}$(Substituting)

$=\frac{-1}{2!}=-\frac{1}{2}$

The second term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is $-\frac{1}{2}$.

The third term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is obtained by substituting $n=2$in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=2$is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^2}{(2\times 2)!}$(Substituting)

$=\frac{1}{4!}=\frac{1}{24}$

The third term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is $\frac{1}{24}$.

The fourth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is obtained by substituting $n=3$in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=3$is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^3}{(2\times 3)!}$( Substituting )

$=\frac{-1}{6!}=\frac{-1}{6\times 5\times 4\times 3\times 2\times 1}=-\frac{1}{720}$

The fourth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is $-\frac{1}{720}$.

The fifth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is obtained by substituting $n=4$in $\frac{(-1{)}^n}{\left(2n\right)!}$. Therefore, the value at $n=4$is:

$\frac{(-1{)}^n}{\left(2n\right)!}=\frac{(-1{)}^4}{(2\times 4)!}$(Substituting )

$=\frac{1}{8!}=\frac{-1}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}=\frac{1}{40320}$

The fifth term of the series $\sum _{n=0}^{\infty }\frac{(-1{)}^n}{\left(2n\right)!}$is $\frac{1}{40320}$.