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I am recently self-studying Complex Analysis and came up with a question with regard to infinite products.

I am trying to show that $\prod_{k=1} {\cos(z/2^k)}$ converges. My first thought is to use the complex representation of cosine, $\sum (-1)^n z^2n/(2n)!$ But I just don't know how to get to the form $\prod (1+b_n)$ for which has a theorem for convergence.

Thanks for all the advice.

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    duplicate of (http://math.stackexchange.com/q/1434117) and (http://math.stackexchange.com/q/706749) – 2017-01-18

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Hint. One may use a Taylor series expansion, observing that as $k \to \infty$, $$ \cos \frac{z}{2^k}=1-\frac{z^2}{2^{2k+1}}+O\left(\frac{z^4}{2^{4k}}\right). $$

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    Have a look at the references I have given. – 2017-01-18
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    @JeanMarie Yes, I was editing my hint to give a second point of view. Thanks. – 2017-01-18
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    @OlivierOloa Thanks Olivier!! Much appreciated! :) – 2017-01-18
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    @Wilson You are welcome. – 2017-01-19