Chapter 11.4, Problem 9OE

To determine

To calculate:

Expanded form of ${\displaystyle \sum _{z=5}^{\infty }\frac{1}{z-1}}$ also read each symbol.

Answer to Problem 9OE

Expanded form of ${\displaystyle \sum _{z=5}^{\infty }\frac{1}{z-1}}$ is $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\mathrm{...}\frac{1}{\infty -1}$ .

Explanation of Solution

Give information:

Expression:

  ${\displaystyle \sum _{z=5}^{\infty }\frac{1}{z-1}}$

Calculation:

It is given that,

  ${\displaystyle \sum _{z=5}^{\infty }\frac{1}{z-1}}$

So here symbol of given expression are:

  $\infty =$ Upper limit of summation.

  $\frac{1}{z-1}=$ Summand of the given expression.(containing first term and last term)

  $5=$ Lower limit of summation.

  $z=$ Index.

Given expression in expanded form is,

  $\begin{array}{l}{\displaystyle \sum _{z=5}^{\infty }\frac{1}{z-1}}=\\ =\frac{1}{5-1}+\frac{1}{6-1}+\frac{1}{7-1}+\frac{1}{8-1}+\mathrm{...}+\frac{1}{\infty -1}\\ =\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\mathrm{...}\frac{1}{\infty -1}\end{array}$

Therefore, expanded form of ${\displaystyle \sum _{z=5}^{\infty }\frac{1}{z-1}}$ is $\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\mathrm{...}\frac{1}{\infty -1}$ .