Question

Determine whether the series $\sum _{k=0}^{\infty }\frac{4^{k+1}}{3^{2k}}$converges or diverges. Give the sum of the convergent series.

Short answer

The series $\sum _{k=0}^{\infty }\frac{4^{k+1}}{3^{2k}}$ converges to $\frac{36}{5}$.

Step-by-step solution

Step 1. Given information.

Given a series $\sum _{k=0}^{\infty }\frac{4^{k+1}}{3^{2k}}$.

Step 2. Find if the series converges or not.

The series$\sum _{k=0}^{\infty }\frac{4^{k+1}}{3^{2k}}$can be expressed as$\sum _{k=0}^{\infty }4{\left(\frac{4}{9}\right)}^k$.

The series $\sum _{k=0}^{\infty }4{\left(\frac{4}{9}\right)}^k$is in the standard form role="math" localid="1648976707514" $\sum _{k=0}^{\infty }c{r}^k$for a geometric series with $c=4$and $r=\frac{4}{9}$.

The geometric series converges if and only if $\left|r\right|<1$.

Since $r=\frac{4}{9}$, it follows that the series role="math" localid="1648976736831" $\sum _{k=0}^{\infty }\frac{4^{k+1}}{3^{2k}}$converges.

Step 3. Find the value to which the series converges.

If the geometric series $\sum _{k=0}^{\infty }c{r}^k$converges, it converges to $\frac{c}{1-r}$.

So, the series $\sum _{k=0}^{\infty }\frac{4^{k+1}}{3^{2k}}$converges to $\frac{4}{1-{\displaystyle \frac{4}{9}}}$, that is $\frac{36}{5}$.