Q. 87 - Calculus

Question

Let $\sum _{k=0}^{\infty }c{r}^k$and $\sum _{k=0}^{\infty }b{v}^k$ be two convergent geometric series. Prove that $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ converges. If neither c nor b is 0, could the series be $\left(\sum _{k=0}^{\infty }c{r}^k\right)\left(\sum _{k=0}^{\infty }b{v}^k\right)$?

Step-by-Step Solution

Answer

$\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)= cb \left(\sum _{k=0}^{\infty }{\left(rv\right)}^k\right)$

Therefore, the expression $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ is also geometric.

Since $\left|rv\right|<1$.

Hence it is proven that $cb\sum _{k=0}^{\infty }{\left(rv\right)}^k or \sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ converges.

And also, $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right) \ne \sum _{k=0}^{\infty }\left(c{r}^k\right).\sum _{k=0}^{\infty }\left(b{v}^k\right)$.

Step 1. Given Information.

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

Step 2. Prove that given expression is geomtric.

When we expand the series $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ we get,

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

As, the above expression is in geometric progression. Therefore the expression $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ is also geometric.

Step 3. Prove that it converges.

Now as we know $\sum _{k=0}^{\infty }\left(c{r}^k\right)$ converges which means $\left|r\right|<1$ and also $\sum _{k=0}^{\infty }\left(b{v}^k\right)$ converges which means $\left|v\right|<1$.

From this, it can be concluded that $\left|rv\right|<1$.

Now, look at the expression from previous step we get $cb\sum _{k=0}^{\infty }{\left(rv\right)}^k$.

And we found that $\left|rv\right|<1$.

So the expression $cb\sum _{k=0}^{\infty }{\left(rv\right)}^k$ or $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right)$ converges.

Step 4. Proof of last part.

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

It is clear from above that the right hand side of the above two equations is not equal therefore the left hand sides also cannot be equal as well

Therefore, $\sum _{k=0}^{\infty }\left(c{r}^k.b{v}^k\right) \ne \sum _{k=0}^{\infty }\left(c{r}^k\right).\sum _{k=0}^{\infty }\left(b{v}^k\right)$.