Question

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

Short answer

The series $\sum _{k=0}^{\infty }\frac{{\left(-3\right)}^{k+1}}{4^{k-2}}$converges to $\frac{-192}{7}$.

Step-by-step solution

Step 1. Given information. 

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

Step 2. Find if the series converges or not.

The series $\sum _{k=0}^{\infty }\frac{{\left(-3\right)}^{k+1}}{4^{k-2}}$can be expressed as $\sum _{k=0}^{\infty }\left(-48\right){\left(\frac{-3}{4}\right)}^k$.

The series role="math" localid="1648977600459" $\sum _{k=0}^{\infty }\frac{{\left(-3\right)}^{k+1}}{4^{k-2}}$is in the standard form $\sum _{k=0}^{\infty }c{r}^k$for a geometric series with $c=-48$and $r=\frac{-3}{4}$.

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

Since$r=\frac{-3}{4}$, it follows that the series converges.

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

If the geometric series $\sum _{k=0}^{\infty }\frac{{\left(-3\right)}^{k+1}}{4^{k-2}}$converges, it converges to $\frac{c}{1-r}$.

So, the series $\sum _{k=0}^{\infty }\frac{{\left(-3\right)}^{k+1}}{4^{k-2}}$converges to $\frac{-48}{1-\left({\displaystyle \frac{-3}{4}}\right)}$, that is $\frac{-192}{7}$.