What is the interval of convergence of #sum_{n=0}^{oo}[log_2(\frac{x+1}{x-2})]^n#? And what's the sum in #x=3#?

1 Answer
MattyMatty
Jan 3, 2018

#]-oo,-4[ " U " ]5, oo[ " is the interval of convergence for x"#
#"x=3 is not in the interval of convergence so sum for x=3 is "oo#

Explanation:

#"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#

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