Question

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

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

Short answer

$\left\{1,\frac{3}{2},\frac{181}{120},\frac{2535}{1680},\frac{228151}{151200}\right\}.$

Step-by-step solution

Step1. Given Information

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

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

Step2. First term

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

Step3. Second term

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

Step4. Third term

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

Step5. Fourth term

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

Step6. Fifth term

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

Step7. Partial sums

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