Question

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

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

Short answer

Ans: The five terms of the series are$1,\frac{1}{2},\frac{1}{120},\frac{1}{1680},\frac{1}{151200}$

Step-by-step solution

Step 1. Given information: 

$\mathbf{\sum }_{\mathbf{k}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{k}\mathbf{!}}{\mathbf{\left(}\mathbf{2}\mathbf{k}\mathbf{\right)}\mathbf{!}}$

Step 2. Finding the first term of the series:

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

$\frac{k!}{\left(2k\right)!}=\frac{0!}{(2\times 0)!}$(Substituting)

$=\frac{0!}{0!}$

=1( Because 0!=1)

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

Step 3. Finding the second term of the series:

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

$\frac{k!}{\left(2k\right)!}=\frac{1!}{(2\times 1)!}$(Substituting)

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

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

Step 4. Finding the third term of the series:

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

$\frac{k!}{\left(2k\right)!}=\frac{3!}{(2\times 3)!}$(Substituting)

$=\frac{3!}{6!}=\frac{3!}{6\times 5\times 4\times 3!}=\frac{1}{120}$

The third term of the series $\sum _{k=0}^{\infty }\frac{k!}{\left(2k\right)!}$is $\frac{1}{120}$.

Step 5. Finding the fourth term of the series:

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

$\frac{k!}{\left(2k\right)!}=\frac{4!}{(2\times 4)!}$(Substituting)

$=\frac{4!}{8!}=\frac{4!}{8\times 7\times 6\times 5\times 4!}=\frac{1}{1680}$

The fourth term of the series $\sum _{k=0}^{\infty }\frac{k!}{\left(2k\right)!}$is $\frac{1}{1680}$.

Step 6. Finding the fifth term of the series:

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

$$\begin{gathered} \frac{k!}{\left(2k\right)!}=\frac{5!}{(2\times 5)!}\left(Substituting\right) \\ =\frac{5!}{10!} \\ =\frac{5!}{10\times 9\times 8\times 7\times 6\times 5!} \\ =\frac{1}{151200} \end{gathered}$$

The fifth term of the series $\sum _{k=0}^{\infty }\frac{k!}{\left(2k\right)!}$is $\frac{1}{151200}$.