Chapter 9.4, Problem 29E

Solution

To calculate : The sum of infinite geometric series if it converges ${\displaystyle \sum _{j=1}^{\infty }3{\left(\frac{1}{4}\right)}^j}$ .

The sum of the series is 1.

Given information:

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

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,

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

The sum of an infinite geometric series is: ${\displaystyle \sum _{j=1}^{\infty }a\cdot r^{j-1}}$

The common ratio ( r ) is: $\frac{1}{4}$

Now equal both the terms: $a\cdot r^{j-1}=3{\left(\frac{1}{4}\right)}^j$

Now find the first term of the sequence using it,

  $\begin{array}{c}{a}_1\cdot r^{1-1}=3{\left(\frac{1}{4}\right)}^1\\ a_1\cdot r^0=3\left(\frac{1}{4}\right)\\ a_1=\frac{3}{4}\end{array}$

For the given series the value of $a_1$ is $\frac{3}{4}$ and $r\text{is}\frac{1}{4}$ . As the common ratio $\left|r\right|<1$ , the given series converges.

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

  $\begin{array}{c}S=\frac{\frac{3}{4}}{1-\frac{1}{4}}\\ =\frac{\frac{3}{4}}{\frac{3}{4}}\\ =1\end{array}$

Thus, the sum of geometric series ${\displaystyle \sum _{j=1}^{\infty }3{\left(\frac{1}{4}\right)}^j}$ is 1.