# How do you find the sum of the infinite geometric series #Sigma 4(1/4)^n# from n=0 to #oo#?

Now subtracting first from second we get

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The sum of the infinite geometric series Σ 4(1/4)^n from n=0 to ∞ is equal to 4/(1 - 1/4), which simplifies to 4/(3/4) = 16/3.

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