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Q 59.

Expert-verifiedFound in: Page 615

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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.

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

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

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.

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" $\frac{\frac{80}{3}}{1-\left(\frac{-1}{4}\right)}$, that is $\frac{64}{3}$.

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