well, you recall from Algebra II that
∞
∑ (4/9)^n = 1/(1 - 4/9) = 9/5
n=0
So just add on the first 4 terms from n = -4
Or, just start with a=(4/9)^-4 and use that in a/(1-r) for the sum
Use the formula for the sum of a geometric series to find the sum or state that the series diverges.
(4/9)^n starting at n=-4
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