Chapter 30, Problem 1RE

To determine

Whether the statement “The series $6+4+\frac{8}{3}+\cdots$ is convergent” is true or false.

Answer to Problem 1RE

The given statement is $\underset{\_}{\text{true}}$.

Explanation of Solution

Result used:

Let ${\displaystyle \sum _{n=0}^{\infty }{a}_1{r}^n}$ be a geometric series of terms, the partial sums $S_n$ represents the sum first n terms of the series, that is $S_n=a_1+a_{1}r+a_1{r}^2+\dots +a_1{r}^{n-1}$.

The series ${\displaystyle \sum _{n=0}^{\infty }{a}_1{r}^n}$ is convergent when $\left|r\right|<1$ and divergent when $\left|r\right|\ge 1$.

Calculation:

The given geometric series is, $6+4+\frac{8}{3}+\cdots$.

The terms of the geometric series can be rewritten as, $6{\left(\frac{2}{3}\right)}^0+6{\left(\frac{2}{3}\right)}^1+6{\left(\frac{2}{3}\right)}^2+\cdots$, that is, ${\displaystyle \sum _{n=0}^{\infty }6{\left(\frac{2}{3}\right)}^n}$.

The above series of the form of ${\displaystyle \sum _{n=0}^{\infty }{a}_1{r}^n}$, where $a_1=6\text{ and }r=\frac{2}{3}$.

Here, the ratio $\left|r\right|<1$, and hence by the above result, the geometric series ${\displaystyle \sum _{n=0}^{\infty }6{\left(\frac{2}{3}\right)}^n}$ is convergent.

Therefore, the given statement is $\underset{\_}{\text{true}}$.