Question

Expand the partial sum and find its value:$\sum _{k=0}^3\frac{2}{k!}$

Short answer

The value of the given partial sum is$\frac{16}{3}.$

Step-by-step solution

Step 1. Given information

$\begin{array}{l}\text{Thepartialsumisgivenas}\\ \sum _{k=0}^3\frac{2}{k!}\\ \text{Wehavetofindthevalueofthepartialsum.}\end{array}$

Step 2. Expand the partial sum

$\begin{array}{l}\text{Now,}\\ \text{Wesubstitutethevalues0,}1,2,3\text{inorder.}\\ \Rightarrow \sum _{k=0}^3\frac{2}{k!}=\frac{2}{0!}+\frac{2}{1!}+\frac{2}{2!}+\frac{2}{3!}\\ =\frac{2}{1}+\frac{2}{1}+\frac{2}{2\times 1}+\frac{2}{3\times 2\times 1}\left[\because n!=n\left(n-1\right)\left(n-2\right)\mathrm{....}\right]\\ =2+2+1+\frac{1}{3}\\ =5+\frac{1}{3}\\ =\frac{\left(5\cdot 3\right)+1}{3}\\ =\frac{16}{3}\end{array}$