Chapter 9.4, Problem 25E

Solution

To calculate : The sum of infinite geometric series if it converges $6+3+\frac{3}{2}+\frac{3}{4}+\cdots$ .

The sum of the series is 12.

Given information:

The given geometric series $6+3+\frac{3}{2}+\frac{3}{4}+\cdots$ .

Formula used: The geometric series ${\displaystyle {\sum }_{k=1}^{\infty }a\cdot r^{k-1}}$ , converges if and only if $\left|r\right|<1$ .If it does converge, the sum: $S=\frac{a_1}{1-r}$ .

Calculation:

Consider the geometric series,

  $6+3+\frac{3}{2}+\frac{3}{4}+\cdots$

The common ratio ( r ) is:

  $\begin{array}{l}r=\frac{3}{6},r=\frac{1}{2}\\ r=\frac{\frac{3}{2}}{3},r=\frac{1}{2}\end{array}$

For the given series the value of $a_1$ is 6 and $r\text{is}\frac{1}{2}$ . As the common ratio $\left|r\right|<1$ , the given series converges.

Now find the sum of infinite geometric series using the formula $S=\frac{a_1}{1-r}$ .

  $\begin{array}{c}S=\frac{6}{1-\frac{1}{2}}\\ =\frac{6}{\frac{1}{2}}\\ =12\end{array}$

Thus, the sum of geometric series $6+3+\frac{3}{2}+\frac{3}{4}+\cdots$ is 12.