Question

A computer program gives the result$\frac{\mathbf{1}}{\mathbf{6}}$ for the sum of the series $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\left(-5\right)}^{\mathbf{n}}$. Show that this series is divergent. Do you see what happened? Warning hint: Always consider whether an answer is reasonable, whether it’s a computer answer or your work by hand.

Short answer

The infinite series$\sum _{n=0}^{\infty }{\left(-5\right)}^n$ is divergent.

Step-by-step solution

Explanation of Solution

Consider Geometric Series $\sum _{n=0}^{\infty }{\left(-5\right)}^n$.

Geometric series

A geometric series is a collection of numbers with a common ratio. This means that each phrase is found by multiplying the term before it.

The formula is provided by,

$\text{G.P}=\frac{a\left(r^n-1\right)}{r-1}$

Here,

a is the first term of series,

r is the common ratio,

n is the total time period.

Calculation

The series is considered divergent since the total does not evaluate to any given value.

$\begin{array}{rcl}\sum _{n=0}^{\infty }{\left(-5\right)}^n& =& {\left(-5\right)}^0+{\left(-5\right)}^1+{\left(-5\right)}^2+{\left(-5\right)}^3+{\left(-5\right)}^4+{\left(-5\right)}^5+....\\ & =& \left(1\right)+\left(-5\right)+\left(25\right)+\left(-125\right)+\left(625\right)+\left(-3125\right)+....\\ & =& ?\\ & & \end{array}$

The infinite summation formula was used by the computer software, which was a mistake.

$\sum _{n=0}^{\infty }c{r}^n=\frac{c}{1-r}$,

Which only applies if $\left|r\right|<1$.

$\begin{array}{rcl}\sum _{n=0}^{\infty }{\left(-5\right)}^n& \ne & \frac{1}{1-\left(-5\right)}\\ & =& \frac{1}{6}\\ & & \end{array}$
Hence, the required result is shown.