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sum(j=0)^(oo)(-3)^(-j)...

`sum_(j=0)^(oo)(-3)^(-j)`

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When the first few terms, `(1)/(1)-(1)/(3)+(1)/(9)-…,` are listed, it can be seen that `t_(1)=1` and the common ratio `r=-(1)/(3)`. Therefore,
`S=(1)/(1-(-(1)/(3)))=(1)/((4)/(3))=(3)/(4)`
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