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So I found this

$$\prod_{k=3}^{\infty} 1 - \tan( \pi/2^k)^4$$

here.

I have only ever done tests for convergence of infinite sums. At this link it shows a way to convert but in this case an is less than one for all k. I can see that $ f(k) = 1 - \tan( \pi/2^k)^4 $ is a strictly increasing function and in fact it is increasing very quickly however it is bound [ $ 1 - \tan( \pi/2^3)^4$, 1) where the lower bound is quite nearly one.

This seems to mean it will converge to something slightly less than the lower bound of the function.

I see that is converges. I want to know if there analytic method to provide a closed from of the solution such as with an alternating series. I figure I can simply do it computationally with MatLab and it would be very simple but I thought there must another way.

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I suppose there is a missing parenthesis right after $2^k$? –  Marek Nov 13 '10 at 23:20
    
Yes, thank you. –  Metahominid Nov 13 '10 at 23:45
1  
Please choose a better title. –  Aryabhata Nov 14 '10 at 0:29

3 Answers 3

up vote 7 down vote accepted

Write $\displaystyle 1- \tan^4 \theta$ as $\dfrac{\cos^2 \theta - \sin^2 \theta}{\cos^4 \theta} = \dfrac{\cos 2\theta}{\cos^4 \theta}$ and make repeated use of the following trick:

To evaluate $\displaystyle \prod_{k=1}^{n}\cos \dfrac{\theta}{2^k}$, multiply and divide by $\sin \dfrac{\theta}{2^n}$ and use $\displaystyle 2\sin \theta \cos \theta = \sin 2\theta$.

This should give you a closed form formula for the product of first $\displaystyle n$ terms, which I believe can be evaluated easily as $\displaystyle n \to \infty$.

For more information on this you can refer: http://en.wikipedia.org/wiki/Vi%C3%A8te's_formula

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@ Moron: Good one! –  user17762 Nov 13 '10 at 23:44
    
@sivaram: Thanks! This is a well known trick, though, and according to the wiki apparently dates back to Euler. I am pretty sure this has been rediscovered a million times since then :-) –  Aryabhata Nov 13 '10 at 23:45
    
I should have remembered my bag of wonderful trigonometry tricks. –  Metahominid Nov 13 '10 at 23:46

First of all I do not understand why you say the product converges to 1. In fact the product must me strictly less than the first term namely, $(1-\tan^4(\frac{\pi}{8}))$. You can however say this function converges since $\displaystyle \sum_{k=1} \tan^4(\frac{\pi}{2^k}) < \infty$. Hence the product converges absolutely.

(Note: $\displaystyle \Pi_{k=1}^n (1+a_k) \leq \exp(\displaystyle \sum_{k=1}^n a_k)$. So convergence of $\displaystyle \sum_{k=1}^{\infty} a_k$ implies the convergence of the infinite product $\displaystyle \Pi_{k=1}^{\infty} (1+a_k)$)

I tried to find an approximate answer using Matlab and then make an educated guess about the limit and then try to prove it. However, the limit of the first $10^7$ terms give me $0.968946146259369$.

I am unable to make an educated guess (something based on $\pi$,$e$ or the golden ratio). It will be interesting to see what this product converges to!

$\textbf{EDIT:}$

This is just for the sake of completion and the solution is due to $\textbf{Moron}$. Following the idea suggested by Moron, and since we know the product has to converge let

$p_n = \displaystyle \Pi_{k=1}^n (1-\tan^4(\frac{\pi}{2^{k+2}})) = \displaystyle \Pi_{k=1}^n \frac{\cos(\frac{\pi}{2^{k+1}})}{\cos^4(\frac{\pi}{2^{k+2}})} = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{1}{\displaystyle \Pi_{k=1}^n \cos^3(\frac{\pi}{2^{k+2}})}$.

$p_n = \frac{\cos(\frac{\pi}{4})}{\cos(\frac{\pi}{2^{n+2}})} \frac{8^n \times \sin^3(\frac{\pi}{2^{n+2}})}{\sin^3(\frac{\pi}{4})}$

Take the limit as $n \rightarrow \infty$.

Hence, $p = \frac{\cos(\frac{\pi}{4})}{1} \times \frac{(\frac{\pi}{4})^3}{\sin^3(\frac{\pi}{4})}$.

Hence, $p = 2 \times (\frac{\pi}{4})^3 = \frac{\pi^3}{32} \approx 0.968946146259369$

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I knew it didn't but I wasn't sure how to articulate convergence to a number very close to one and in retrospect I see must converge to strictly less the lower bound of the function. –  Metahominid Nov 13 '10 at 23:35
    
There is a more detailed discussion on convergence issues of infinite products, in the book "Complex Analysis" by Ahlfors. –  user1119 Nov 13 '10 at 23:41
    
Thank you, awesome. I will attempt to get it, but I am also taking the course next semester. –  Metahominid Nov 14 '10 at 2:21

Convert the product to a sum by taking logarithms. For $x$ small, $\tan x \approx x$ and $\log (1 - x) \approx -x$ and so after taking logs, your product is roughly like the convergent geometric series $\sum \pi 2^{-k}$. So it converges.

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