Chapter 11.4, Problem 21WE

To determine

To calculate: The sigma notation

Answer to Problem 21WE

The sigma notation is ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}$

Explanation of Solution

Given information: $-9+3-1+\frac{1}{3}+\mathrm{........}$

Let, numbers of terms are $n$ , where $n=1,2,3,4,\mathrm{...............}\infty$

The sigma notation of the series $-9+3-1+\frac{1}{3}+\mathrm{........}$

  $\therefore {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}$

Verify, the series

Put, $n=1,2,3,4,\mathrm{...............}\infty$

Calculation:

  $\begin{array}{l}\therefore {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}={\left(-1\right)}^1{\left(9\right)}^1\frac{1}{3^{2-2}}+{\left(-1\right)}^2{\left(9\right)}^2\frac{1}{3^{4-2}}+{\left(-1\right)}^3{\left(9\right)}^3\frac{1}{3^{6-2}}+\mathrm{..............}\infty \\ \Rightarrow {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}=-9+\frac{27}{9}-\frac{81}{81}+\mathrm{........}\infty \\ \Rightarrow {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}=-9+\frac{27}{9}-\frac{81}{81}+\mathrm{........}\infty \\ \Rightarrow {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}=-9+3-1+\mathrm{..........}\infty \end{array}$

Hence, ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n{\left(9\right)}^n\frac{1}{3^{2n-2}}}$ is the sigma notation of the $-9+3-1+\frac{1}{3}+\mathrm{........}$ .