Provide the first five terms of the sequence of partial sums for the given series.

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

$\text{Consider the geometric series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{.}$

The objective is to provide the first five terms of partial sums for the given series.

The strategy to find the first five terms of partial sums for the given series is to find the first five terms of the series$\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}$.

$$\begin{gathered} \text{The first term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is obtained by substituting}k=0\text{in}\frac{k!}{\left(2k\right)!}\text{.} \\ \text{Therefore, the value at}i=0\text{is:} \\ \begin{array}{l}\frac{k!}{\left(2k\right)!}=\frac{0!}{(2\times 0)!}\text{(Substituting)}\\ =\frac{0!}{0!}\\ =1(\text{Because}0!=1)\\ \text{The first term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is}1\text{.}\end{array} \end{gathered}$$

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

$$\begin{gathered} \text{The third term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is obtained by substituting}k=2\text{in}\frac{k!}{\left(2k\right)!}\text{.} \\ \text{Therefore, the value at}k=2\text{is:} \\ \begin{array}{l}\frac{k!}{\left(2k\right)!}=\frac{3!}{(2\times 3)!}\text{(Substituting)}\\ =\frac{3!}{6!}\\ =\frac{3!}{6\times 5\times 4\times 3!}\\ =\frac{1}{120}\\ \text{The third term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is}\frac{1}{120}\text{.}\end{array} \end{gathered}$$

$$\begin{gathered} \text{The fourth term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is obtained by substituting}k=3\text{in}\frac{k!}{\left(2k\right)!}\text{.} \\ \text{Therefore, the value at}k=3\text{is:} \\ \begin{array}{l}\frac{k!}{\left(2k\right)!}=\frac{4!}{(2\times 4)!}\text{(Substituting)}\\ =\frac{4!}{8!}\\ =\frac{4!}{8\times 7\times 6\times 5\times 4!}\\ =\frac{1}{1680}\\ \text{The fourth term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is}\frac{1}{1680}\text{.}\end{array} \end{gathered}$$

$$\begin{gathered} \text{The fifth term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is obtained by substituting}k=4\text{in}\frac{k!}{\left(2k\right)!}\text{.} \\ \text{Therefore, the value at}k=4\text{is:} \\ \begin{array}{l}\frac{k!}{\left(2k\right)!}=\frac{5!}{(2\times 5)!}\text{(Substituting)}\\ =\frac{5!}{10!}\\ =\frac{5!}{10\times 9\times 8\times 7\times 6\times 5!}\\ =\frac{1}{151200}\\ \text{The fifth term of the series}\sum _{k=0}^{\mathrm{\infty }} \frac{k!}{\left(2k\right)!}\text{is}\frac{1}{151200}\text{.}\end{array} \end{gathered}$$

$$\begin{gathered} \text{The first five terms in the sequence of partial sums are:} \\ \begin{array}{l}{S}_1=1\\ S_2=S_1+a_2\\ =1+\frac{1}{2}\text{(Substitution)}\\ =\frac{3}{2}\\ S_3=S_2+a_3\\ =\frac{3}{2}+\frac{1}{120}\text{(Substitution)}\\ =\frac{180+1}{120}\\ =\frac{181}{120}\\ \begin{array}{l}{S}_4=S_3+a_4\\ =\frac{181}{120}+\frac{1}{1680}\text{(Substitution)}\\ =\frac{2534+1}{1680}\\ =\frac{2535}{1680}\\ S_5=S_4+a_5\\ =\frac{2535}{1680}+\frac{1}{151200}\text{(Substitution)}\\ =\frac{228150+1}{151200}\\ =\frac{228151}{151200}\\ \text{Therefore, first five terms of partial sums for the given series is}\left\{1,\frac{3}{2},\frac{181}{120},\frac{2535}{1680},\frac{228151}{151200}\right\}\text{.}\end{array}\end{array} \end{gathered}$$