Chapter 12, Problem 48RE

To determine

To find: Whether the geometric series converges or diverges. Find the sum if the series converges.

Answer to Problem 48RE

The infinite series converges and its sum is $\frac{12}{7}$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{k=1}^{\infty }3{\left(-\frac{3}{4}\right)}^{k-1}}$ .

Calculation:

Consider the formula for the geometric series is,

  ${\displaystyle \sum _{k=1}^{\infty }{a}_1{r}^{k-1}}=\frac{a_1}{1-r}$

Consider given series is,

  ${\displaystyle \sum _{k=1}^{\infty }3{\left(-\frac{3}{4}\right)}^{k-1}}$

The above series is convergent when $\left|r\right|<1$ as $\left|-\frac{3}{4}\right|<1$ is true the series converges.

Then, the sum is,

  $\begin{array}{c}{\displaystyle \sum _{k=1}^{\infty }3{\left(-\frac{3}{4}\right)}^{k-1}}=\frac{3}{1-\left(-\frac{3}{4}\right)}\\ =\frac{3}{1+\frac{3}{4}}\\ =\frac{12}{7}\end{array}$

Thus, the infinite series converges and its sum is $\frac{12}{7}$ .