Question

Determine whether the series $\frac{80}{3}-\frac{20}{3}+\frac{5}{3}-\frac{5}{12}+\dots$converges or diverges. Give the sum of the convergent series.

Short answer

The series $\frac{80}{3}-\frac{20}{3}+\frac{5}{3}-\frac{5}{12}+\dots$converges to$\frac{64}{3}$.

Step-by-step solution

Step 1. Given information.

Given a series $\frac{80}{3}-\frac{20}{3}+\frac{5}{3}-\frac{5}{12}+\dots$.

Step 2. Find if the series converges or not.

The standard form of a geometric series is $\sum _{k=0}^{\infty }c{r}^k$.

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

In the series $\frac{80}{3}-\frac{20}{3}+\frac{5}{3}-\frac{5}{12}+\dots$it can be seen that $c=\frac{80}{3}$.

Every term after that is $-\frac{1}{4}$times the previous term.

It follows that $r=-\frac{1}{4}$.

Since $r=-\frac{1}{4}$, the series$\frac{80}{3}-\frac{20}{3}+\frac{5}{3}-\frac{5}{12}+\dots$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 localid="1648982757515" $\frac{80}{3}-\frac{20}{3}+\frac{5}{3}-\frac{5}{12}+\dots$converges to localid="1648982761274" $\frac{{\displaystyle \frac{80}{3}}}{1-\left({\displaystyle \frac{-1}{4}}\right)}$, that is $\frac{64}{3}$.