How do you find the sum of the infinite geometric series $Sigma 2(2/3)^n$ from n=0 to $oo$ ?

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Answer 1 · 2018-04-27T04:01:15

$sum2(2/3)^n = 6$

$sum2(2/3)^n$ for n=0 to infinity

=2 $(sum(2/3)^n)$

Sum of an infinite geometric series (where the ratio r is $-1<=r<=1$ ) is given by $a/(1-r)$ where r is the common ratio and ‘a’ is the first term.

Therefore,
$sum(2/3)^n = 1/(1-(2/3))$

$=1/(1/3) = 3$

Substituting this value in our original equation gives us -

2 $(sum(2/3)^n) = 2*3 = 6$

How do you find the sum of the infinite geometric series Sigma 2(2/3)^nSigma 2(2/3)^n from n=0 to oooo?