Find the sum of the infinite geometric series: 9 – 6 + 4 – …

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The defining characteristic of all geometric sequences is a common ratio which is a constant when dividing any term by the term preceding it. In this case the common ratio is:  -6/9=4/-6=r=-2/3 An infinite series will have a sum when r^2<1, so in this case the sum will converge to an actual value because (-2/3)^(+oo) approaches zero. The sum of any geometric sequence is: s(n)=a(1-r^n)/(1-r), since we have a common ratio of -2/3 and we want to calculate an infinite series, ie, n approaches infinity, the sum becomes simply: s(n)=a/(1-r)    (because (1-r^+oo) approaches 1 as n approaches +oo) So our infinite sum is: s(+oo)=9/(1--2/3) s(+oo)=9/(1+2/3) s(+oo)=9/(5/3) s(+oo)=27/5 s(+oo)=54/10 s(+oo)=5.4

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