show me the steps please. Created by h. science-mathematics-en - mathematics-en

determine whether the infinite geometric series is convergent or divergent. if its convergent find its sum (1/(√3))+(1/3)+(1/(3√3))+1/9+...?

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determine whether the infinite geometric series is convergent or divergent. if its convergent find its sum (1/(√3))+(1/3)+(1/(3√3))+1/9+...?

show me the steps please

Since 1/sqrt(3) is between -1 and 1, then it will be convergent

S = (1/sqrt(3)) + (1/3) + (1/(3sqrt(3))) + (1/9) + ...

S * (1/sqrt(3)) = (1/3) + (1/(3sqrt(3))) + (1/9) + ....

S - S * (1/sqrt(3)) = (1/sqrt(3)) + (1/3) - (1/3) + (1/(3sqrt(3))) - (1/(3sqrt(3))) + (1/9) - (1/9) + ....

S * (1 - 1/sqrt(3)) = 1/sqrt(3) + 0 + 0 + 0 + ...

S * ((sqrt(3) - 1) / sqrt(3)) = 1 / sqrt(3)

S * (sqrt(3) - 1) = 1

S = 1 / (sqrt(3) - 1)

S = (sqrt(3) + 1) / (3 - 1)

S = (sqrt(3) + 1) / 2

Its common ratio is 1/3 ÷ 1/√3 = √3 / 3 which is between 0 and 1 so it's converging. The rule for sum is a1 / (1 - r)

so (1/√3) / (1 - √3 / 3)

= √3 / 3 ÷ (3/3 - √3/3)

= √3/3 ÷ ((3 - √3)/3)

= √3/3 • 3/(3 - √3)

= √3 / (3 - √3)