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Answer 1 · 2017-02-26T00:30:17

Radius of convergence: $abs ( x ) < 2$

$(x-1)/(x+2)$

$= (x + 2 - 3)/(x+2)$

$= 1 - (3)/(x+2)$

$= 1 - (3/2)/(x/2+1)$

$= 1 - 3/2(1+x/2)^(-1)$

From here we can use the Binomial/Taylor Expansion as follows:

$= 1 - 3/2(1 +( -1) x/2 + (((-1)(-2))/(2!)) (x/2)^2 + (((-1)(-2)(-3))/(3!)) (x/2)^3 + mathcal O (x)^4)$

$= 1 - 3/2(1 - x/2 + x^2/4 - x^3 /8 + mathcal O (x)^4)$

$= -1/2 + 3/4 x - 3/8 x ^2 + 3/16 x^3 + mathcal O (x)^3$

I admit this is slack but there is such an obvious pattern there. So if we apply the ratio test to the second and third term, we see that:

$abs ((- 3/8 x ^2)/(3/4 x) )= abs (( x )/(2 ) )$

This series absolutely converges if: $abs (( x )/(2 ) ) < 1$

ie if $abs ( x ) < 2$