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Determine the values of $x$ for which the infinite product $\prod_{n=0}^{\infty} \left ( 1 + \frac{1}{x^{2n}} \right )$ converges absolutely. Find its value whenever it converges.
asked Sep 9, 2017 by randomisation (1,920 points)  

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Let $y = \prod_{n=0}^{\infty} \left ( 1 + \frac{1}{x^{2n}} \right ) \implies log(y) = \sum_{n=0}^{\infty} log \left ( 1 + \frac{1}{x^{2n}} \right )$

Now, let $f_n(x) = log \left ( 1 + \frac{1}{x^{2n}} \right ) $

Using ratio test,

For $0<|x|\leq1, \frac{f_{n_1}(x)}{f_n(x)} \geq 1$, and thus it's divergent (Equality holds for $x=1$)

For $1<|x|<\infty, \frac{f_{n_1}(x)}{f_n(x)} < 1$, and hence the series $ log(y) $ is convergent

Now, the value of $log(y)$ is greater than 0, so we can take anti-log,

$\implies$ the given series  $\prod_{n=0}^{\infty} \left ( 1 + \frac{1}{x^{2n}} \right )$ is also convergent for $1<|x|<\infty$
answered Sep 9, 2017 by randomisation (1,920 points)  
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