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1)1 + 2x + 4x² + 8x³ + ...The 1st term = 1
Common ratio = 2x- 1 < 2x < 1 for it converges.- 1/2 < x < 1/2The sum to ∞= 1 / (1 - 2x) 2)(a+x) - 1 + 1/(a+x) - 1/(a+x)² + ...The 1st term = (a+x) - 1Common ratio = 1/(a+x)² - 1 < 1/(a+x)² < 1 for it converges.Since 0 < 1/(a+x)² ,1/(a+x)² < 11 < (x+a)²(x+a-1)(x+a+1) > 0(x - (1-a)) (x - (1+a)) > 0x < 1-a or x > 1+a for a ≥ 0x < 1+a or x > 1-a for a ≤ 0The sum to ∞ = (a+x) / (1 - 1/(a+x)²)= (a+x) / ((a+x)² - 1)/(a+x)²)= (a+x)³ / [(a+x-1)(a+x+1)]


2011-03-27 16:49:42 補充：
Because of the square of any real number >= 0 ,

so 0 < (1/(a+x))² = 1/(a+x)² is always true for any real x.

We just need to solve 1/(a+x)² < 1.

2011-03-27 16:50:34 補充：
(x+a)² - 1 = (x+a-1)(x+a+1)

Q2)
I don't understand the part starting from "Since 0 < 1/(a+x)² " until the end.
can u briefly explain...these??
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