Question

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

Short answer

The value of the partial sum$\sum _{k=0}^3\frac{3}{k!}$is 8.

Step-by-step solution

Step 1. Given information

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

Step 2. Expand the partial sum

$\begin{array}{l}\text{Now,}\\ \text{Wesubstitutethevalues0,}1,2,3\text{inorder.}\\ \Rightarrow \sum _{k=0}^3\frac{3}{k!}=\frac{3}{0!}+\frac{3}{1!}+\frac{3}{2!}+\frac{3}{3!}\\ =\frac{3}{1}+\frac{3}{1}+\frac{3}{2\times 1}+\frac{3}{3\times 2\times 1}\left[\because n!=n\left(n-1\right)\left(n-2\right)\mathrm{....}\right]\\ =3+3+\frac{3}{2}+\frac{1}{2}\\ =6+\frac{4}{2}\\ =6+2\\ =8\end{array}$