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Please help answer this geometric question!!

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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For what values of x, \[x \neq-1\] will the following infinite geometric series have a finite sum? \[(x+1) + (x+2)^2 + (x+1)^3 ....\]
when -1 < x+1 < 0 --> -2 < x < -1 and 0< x+2 < 1 --> -2 < x < -1 this way the series converges and the sum will be finite
pls explain properly

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no it's infinite 'cause the exponents are growing up
another way of looking at it is to separate it into 2 series and use the formula for sum of a geometric series Sn = a(1 - r^n)/(1-r) For series (x+1)^n , n is odd going to infinity, r = (x+1)^2 S = (x+1)(1-(x+1)^2n) / (1-(x+1)^2) = [x+1 -(x+1)^2n+1] / -(x^2+2x) For S to not be +-infinity, you need (x+1)^n to go to zero thus 0<|x+1| <1 S = (x+1)/-(x^2 +2x) For series (x+2)^n , n is even going to infinity, r = (x+2)^2 S = (x+2)^2(1-(x+2)^2n) / (1-(x+2)^2) = [(x+2)^2 -(x+1)^2n+2] / -(x^2 +4x+3) For S to not be +-infinity, you need (x+2)^n to go to zero thus 0<|x+2| <1 S = (x+2)^2/-(x^2 +4x +3) So Basically we need to find an x such that 0 < |x+1| < 1 and 0<|x+2| < 1 see prev post -2 < x < -1

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