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

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

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Answer 1 · 2018-01-03T09:52:32

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

Explanation:

$Treat the sum as would it be a geometric series by substituting$ $"Treat the sum as would it be a geometric series by substituting"$
$z = {log}_{2} (\frac{x + 1}{x - 2})$ $"z = log_2((x+1)/(x-2))$
$Then we have$ $"Then we have"$
$\sum n = 0 z^{n} = \frac{1}{1 - z} for | z | < 1$ $sum_{n=0} z^n = 1/(1-z) " for " |z| < 1$
$So the interval of convergence is$ $"So the interval of convergence is"$
$- 1 < {log}_{2} (\frac{x + 1}{x - 2}) < 1$ $-1 < log_2((x+1)/(x-2)) < 1$
$\Rightarrow \frac{1}{2} < \frac{x + 1}{x - 2} < 2$ $=> 1/2 < (x+1)/(x-2) < 2$
$\Rightarrow \frac{x - 2}{2} < x + 1 < 2 (x - 2) OR$ $=> (x-2)/2 < x+1 < 2(x-2) " OR "$
$\frac{x - 2}{2} > x + 1 > 2 (x - 2) (x-2 negative)$ $(x-2)/2 > x+1 > 2(x-2) "(x-2 negative)"$

$Positive case :$ $"Positive case : "$
$\Rightarrow x - 2 < 2 x + 2 < 4 (x - 2)$ $=> x-2 < 2x + 2 < 4(x-2)$
$\Rightarrow 0 < x + 4 < 3 (x - 2)$ $=> 0 < x + 4 < 3(x-2)$
$\Rightarrow - 4 < x < 3 x - 10$ $=> -4 < x < 3x-10$
$\Rightarrow x > - 4 and x > 5$ $=>x > -4 and x > 5$
$\Rightarrow x > 5$ $=> x > 5$

$Negative case :$ $"Negative case : "$
$- 4 > x > 3 x - 10$ $-4 > x > 3x-10$
$\Rightarrow x < - 4 and x < 5$ $=> x < -4 and x < 5$
$\Rightarrow x < - 4$ $=> x < -4$

$second part : x = 3 \Rightarrow z = 2 > 1 \Rightarrow sum is \infty$ $"Second part : "x=3 => z = 2 > 1 => "sum is "oo$

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

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