Evaluation of $\lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{1}{1-\tan^2(2^{-n})}$

Asked May 17 '16 at 12:10

Active May 17 '16 at 12:18

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Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{1}{1-\tan^2(2^{-n})}$

$\bf{My\; Try::}$ Let $p_{k} = \displaystyle \frac{1}{1-\tan^2(2^{-n})} = \frac{\cos^2(2^{-k})}{\cos(2^{-k+1})}$

So $\displaystyle p_{1}p_{2}p_{3}...........p_{n} = \frac{\cos^2(2^{-1})}{\cos(2^{0})}\cdot \frac{\cos^2(2^{-2})}{\cos(2^{-1})}\cdot \frac{\cos^2(2^{-3})}{\cos(2^{-2})}\cdot.........\frac{\cos^2(2^{-n})}{\cos^2(2^{-n+1})}$

Now How can I solve after that, Help required, Thanks

edited May 17 '16 at 12:18

Simply Beautiful Art

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asked May 17 '16 at 12:10

juantheron

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Which is correct $1/\tan^2(2^{-n})$ (in the title) or $1/(1-\tan^2(2^{-n}))$ (in the body)? – mathlove May 17 '16 at 12:16
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You can obviously do some cancelling. – Simply Beautiful Art May 17 '16 at 12:17
See http://math.stackexchange.com/questions/455995/finding-the-limit-lim-limits-n-to-infty-cos-frac-x-2-cdot-cos-frac – lab bhattacharjee May 17 '16 at 12:17
@mathlove I think its $1-\tan^2(2^{-n})$ or else $\lim_{n\to\infty}a_n=undefined$ – Simply Beautiful Art May 17 '16 at 12:17
In addition to telescoping sum, there is something called telescoping product. this is one example of it. – achille hui May 17 '16 at 12:41
You mean $\prod\limits_{k=1}^\infty \frac{1}{1-\tan^2(2^{-k})}$ ? Then look at http://math.stackexchange.com/questions/455995/finding-the-limit-lim-limits-n-to-infty-cos-frac-x-2-cdot-cos-frac?lq=1 . – user90369 May 17 '16 at 12:51

Evaluation of $\lim_{n\rightarrow \infty}\prod^{n}_{k=1}\frac{1}{1-\tan^2(2^{-n})}$

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