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Determine whether the geometric series is convergent or divergent. And if it is convergent, what is the sum?
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\[\sum_{n=1}^{\infty} \frac{ (-3)^{n-1} }{ 4^{n} }\]
i just don't know which value a and r is.
what are our criteria for convergence? and the r will be the part that his the exponent
\[\frac{(-3)^{n-1}}{4^n}\] \[\frac{(-1)^{n-1}~{3^n}~3^{-1}}{4^n}\] etc ...
you kinda of lost me here.
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i know for convergence, r needs to be less than 1
convergence is when r is between -1 and 1
yeah, but im not sure how to figure ou how to find r.
well let c n1= 1/4 right?
?
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