Chapter 11.4, Problem 67HP

To determine

To explain why an arithmetic series is always divergent.

Answer to Problem 67HP

An arithmetic series is always divergent

Explanation of Solution

Given:

An infinite arithmetic series.

Concept Used:

Sum of first n terms of an arithmetic series whose first term is a and common difference d is:

  $S_n=\frac{n\left(2a+\left(n-1\right)d\right)}{2}$

Calculation:

To explain why an arithmetic series is always divergent.

First consider an infinite arithmetic series

  $a+\left(a+d\right)+\left(a+2d\right)+\left(a+3d\right)\mathrm{......}={\displaystyle {\sum }_{n=0}^{\infty }\left(a+nd\right)}$ .

Then sum of its n term is given by the formula:

  $\begin{array}{c}{S}_n=\frac{n\left(2a+\left(n-1\right)d\right)}{2}\\ =\left(a-\frac{d}{2}\right)n+\frac{d}{2}{n}^2\end{array}$

Note that both the terms of above sum tends to infinity as n tends to infinity, i.e.

  $\begin{array}{l}\left(a-\frac{d}{2}\right)n\to \infty ;\text{ and }\frac{d}{2}{n}^2\to \infty \\ \text{When }n\to \infty \end{array}$

Thus, the above sum diverges a n tends to infinity, since

  $\begin{array}{c}{S}_n=\left(a-\frac{d}{2}\right)n+\frac{d}{2}{n}^2\\ =\infty +\infty \text{ when }n=\infty \end{array}$

Thus, an arithmetic series is always divergent.