Chapter 8, Problem 4P

To determine

To find: The sum of the series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$ .

Answer to Problem 4P

The sum of the given series is $\frac{17}{6}$ .

Explanation of Solution

Given:

The series is $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$ .

Calculation:

The given series can be divided into three series as follows.

  $\begin{array}{c}S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots \\ =1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\mathrm{...}+\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{4}+\mathrm{...}\right)+\frac{1}{9}\left(1+\frac{1}{3}+\mathrm{...}\right)\\ =S_1+S_2+S_3\end{array}$

Find the sum of series $S_1$ .

  $S_1=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\mathrm{...}$

This series is in geometric progression and the sum of $n$ terms of the G.P. is given by $S_n=\frac{a}{1-r}$ .

Substitute $1$ for $a$ , and $\frac{1}{2}$ for $r$ in the equation $S_n=\frac{a}{1-r}$ .

  $\begin{array}{c}{S}_1=\frac{1}{1-\frac{1}{2}}\\ =2\end{array}$

Find the sum of series $S_2$ .

  $S_2=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\mathrm{...}\right)$

This series is in geometric progression and the sum of $n$ terms of the G.P. is given by $S_n=\frac{a}{1-r}$ .

Substitute $1$ for $a$ , and $\frac{1}{2}$ for $r$ in the equation $S_n=\frac{a}{1-r}$ .

  $\begin{array}{c}{S}_2=\frac{1}{3}\left(\frac{1}{1-\frac{1}{2}}\right)\\ =\frac{2}{3}\end{array}$

Find the sum of series $S_3$ .

  $S_3=\frac{1}{9}\left(1+\frac{1}{3}+\mathrm{...}\right)$

This series is in geometric progression and the sum of $n$ terms of the G.P. is given by $S_n=\frac{a}{1-r}$ .

Substitute $1$ for $a$ , and $\frac{1}{3}$ for $r$ in the equation $S_n=\frac{a}{1-r}$ .

  $\begin{array}{c}{S}_3=\frac{1}{9}\left(\frac{1}{1-\frac{1}{3}}\right)\\ =\frac{1}{9}\left(\frac{3}{2}\right)\\ =\frac{1}{6}\end{array}$

Find the sum of the given series.

  $\begin{array}{c}S=S_1+S_2+S_3\\ =2+\frac{2}{3}+\frac{1}{6}\\ =\frac{17}{6}\end{array}$

Therefore, the sum of the given series is $\frac{17}{6}$ .