# calculus

determine whether the series is convergent if so find sum

it is the sum from k=1 to infinity of ((-1)^k)/(3^(k+1))
i found this series to be geometric where
a=-1/9 and
r=1/3
my answer was converges to 1/6

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1. (-1)^k
---------
3^(k+1)

-1/9 , 1/27 , -1/3^4 ...

yes, g = -1/9
no r = -1/3 (disagree with your sign)

Sum = -(1/9 ) / (1 +1/3) =-1/12

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