Question

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`?

Answer 1

`]-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`