Q. 89 - Calculus

Question

Let$\sum _{k=0}^{\infty }c{r}^k$and $\sum _{k=0}^{\infty }b{v}^k$ be two convergent geometric series. If b and v are both nonzero, prove that  $\sum _{k=0}^{\infty }\frac{c{r}^k}{b{v}^k}$  is a geometric series. What condition(s) must be met for this series to converge?

Step-by-Step Solution

Answer

$\sum _{k=0}^{\infty }\frac{c{r}^k}{b{v}^k} = \frac{c}{b}\sum _{k=0}^{\infty }{\left(\frac{r}{v}\right)}^k$ and so the series is in geometric progression.

The condition for the series $\sum _{k=0}^{\infty }\frac{c{r}^k}{b{v}^k}$ to be convergent is $\left|\frac{r}{v}\right|<1$.

Step 1. Given Information.

$\sum _{k=0}^{\infty }c{r}^k$ and $\sum _{k=0}^{\infty }b{v}^k$ be two convergent series.

Step 2. Prove that the series is geometric.

The expanded form of the series $\sum _{k=0}^{\infty }\frac{c{r}^k}{b{v}^k}$ will be:

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{c{r}^k}{b{v}^k}=\frac{c}{b}\left(1+\frac{r}{v}+{\left(\frac{r}{v}\right)}^2+...\right) \\ =\frac{c}{b}\sum _{k=0}^{\infty }\frac{r^k}{v^k} \\ =\frac{c}{b}\sum _{k=0}^{\infty }{\left(\frac{r}{v}\right)}^k \end{gathered}$$

Therefore the series is in geometric progression.

Step 3. Conditions met for the series to converge.

For any geometric series to converge, its common ratio must be less than 1.

The series $\sum _{k=0}^{\infty }\frac{c{r}^k}{b{v}^k}$ is in geometric progression and its common ratio is $\frac{r}{v}$.

So the condition for the series to converge is given as $\left|\frac{r}{v}\right|<1$.