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Determine whether the infinite geometric series is convergent or divergent. If it is convergent, find its sum. (If the series is divergent, enter DIVERGENT.)

(1)-(1/6)+(1/36)-(1/216)+... ...?

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The geometric series can be rewritten as
1 + (-1/6) + (-1/6)^2 + (-1/6)^3 + ...
Since the magnitude of the common ratio is less than 1. The series is then convergent and it has a sum.
S = a/(1-r) = 1/(1+1/6) = 6/7

I hope my answer has come to your help.
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