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Find the sum of the following infinite geometric series, if it exists. 1/2, 1/4, 1/8, 1/16,...
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Well first the equation you will use is
\[S _{n}=\frac{ a(1-r^n) }{ 1-r }\]
\[a=\frac{ 1 }{ 2 }\] \[r=\frac{ 1 }{ 2 }\]
Find \[S_{n}\]
\[a _{1}/1-r ^{n}\] if \[\left| r \right|<1\] so there fore like aztec said r and a are 1/2 and 1/2
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Then just substitute from there
Oh crap, it said infinite. @ronaldo7
\[S _{\infty}=\frac{ a(1-r^\infty) }{ 1-r }\] When \[\left| r \right|<1\] And it is raised by the power of infinity, it will become so small and be close to zero. That means r^infinity is zero. \[S _{\infty}=\frac{ a(1-0) }{ 1-r }\] \[S _{\infty}=\frac{ a }{ 1-r }\]
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