How do I find the sum of the infinite geometric series such that $a_{1} = - 5$ $a_1=-5$ and $r = \frac{1}{6}$ $r=1/6$ ?

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How do I find the sum of the infinite geometric series such that $a_{1} = - 5$ $a_1=-5$ and $r = \frac{1}{6}$ $r=1/6$ ?

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Answer 1 · 2015-05-23T22:03:39

For the infinite geometric series $S = a_{1} + a_{1} r + a_{1} r^{2} + \dots$ $S=a_1+a_1r+a_1r^2+cdots$ we can use the formula $S = \frac{a_{1}}{1 - r}$ $S=a_1/(1-r)$ as long as $| r | < 1$ $|r|<1$ .

In our case, $a_{1} = - 5, r = \frac{1}{6}$ $a_1=-5, r=1/6$ and $| r | = ∣ ∣ ∣ \frac{1}{6} ∣ ∣ ∣ = \frac{1}{6} < 1$ $|r|=|1/6|=1/6<1$ so
$S = \frac{- 5}{1 - \frac{1}{6}} = \frac{- 5}{\frac{5}{6}} = - 5 \cdot \frac{6}{5} = - 6$ $S=(-5)/(1-1/6)=(-5)/(5/6)=-5* 6/5=-6$

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How do I find the sum of the infinite geometric series such that $a_{1} = - 5$ $a_1=-5$ and $r = \frac{1}{6}$ $r=1/6$ ?

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