Q 57. - Calculus

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.

Step-by-Step Solution

Answer

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

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