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1/(1-x) = Sigma x^n for |x|<1 alright. It's said that the derivative of 1/(1-x) = Sigma x^n is equal to: 1/(1-x)^2 = Sigma n x^(n-1) yet should not there be a minus sign before the Sigma?? ((x-1)^-1)' = (-1) (x-1)^(-2) = Sigma n x^(n-1) (x-1)^(-2) = (-1) Sigma n x^(n-1)
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Pudding is all you need to solve it.
You're correct! the result is negative!
I forgot to differentiate the 1-x itself ((1-x)^(-1))' =[(-1) (1-x)^(-2)] [-1] = Sigma n x^(n-1) //[-1] is the derivative of 1-x (1-x)^(-2) = Sigma n x^(n-1) fix'd
can anyone tell me, why this post is poping up now? its 3 years old :D
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