"Treat the sum as would it be a geometric series by substituting"
"z = log_2((x+1)/(x-2))
"Then we have"
sum_{n=0} z^n = 1/(1-z) " for " |z| < 1
"So the interval of convergence is"
-1 < log_2((x+1)/(x-2)) < 1
=> 1/2 < (x+1)/(x-2) < 2
=> (x-2)/2 < x+1 < 2(x-2) " OR "
(x-2)/2 > x+1 > 2(x-2) "(x-2 negative)"
"Positive case : "
=> x-2 < 2x + 2 < 4(x-2)
=> 0 < x + 4 < 3(x-2)
=> -4 < x < 3x-10
=>x > -4 and x > 5
=> x > 5
"Negative case : "
-4 > x > 3x-10
=> x < -4 and x < 5
=> x < -4
"Second part : "x=3 => z = 2 > 1 => "sum is "oo