8
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Given this double infinite products

$$\prod_{k=1}^{\infty}\left(\prod_{j=1}^{\infty}{(2j)^{2^k}-1\over (2j)^{2^k}+1}\right)^{1/2^k}={2\over \pi}\tag1$$

How can we prove $(1)$?

An attempt:

Take the log

$$\sum_{k=1}^{\infty}{1\over 2^k}\sum_{j=1}^{\infty}\ln\left({(2j)^{2^k}-1\over (2j)^{2^k}+1}\right)=\log{2\over \pi}\tag2$$

Not sure how to handle $$\sum_{j=1}^{\infty}\ln\left({(2j)^{2^k}-1\over (2j)^{2^k}+1}\right)\tag3$$

Recall $(4)$, maybe it could be of some use for $(3)$

$$\ln{1+x\over 1-x}=\sum_{n=1}^{\infty}{2\over (2n-1)}x^{2n-1}\tag4$$

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  • $\begingroup$ Indeed, what happened when you used (4) (for $|x|<1$) in each term of the series in (3)? $\endgroup$ – Did Mar 20 '17 at 13:20
  • $\begingroup$ Hint : $${(2j)^{2k}-1\over(2j)^{2k}+1} = 1-{2\over (2j)^{2k}+1}$$ $\endgroup$ – Furrane Mar 20 '17 at 13:23
  • $\begingroup$ @Furrane Really? Are you able to follow on your hint? $\endgroup$ – Did Mar 20 '17 at 13:26
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$$\sum_{j\geq 1}\log\left(\frac{(2j)^{2^k}-1}{(2j)^{2^k}+1}\right) = \sum_{j\geq 1}\log\left(\frac{1-\frac{1}{(2j)^{2^k}}}{1+\frac{1}{(2j)^{2^k}}}\right)=-\sum_{j\geq 1}\sum_{m\geq 0}\frac{2}{(2m+1)(2j)^{2^k(2m+1)}}\tag{1}$$ by multiplying both sides of $(1)$ by $\frac{1}{2^k}$ and summing over $k\geq 1$ we are left with

$$ -\sum_{j\geq 1}\sum_{m\geq 0}\sum_{k\geq 1}\frac{2}{2^k(2m+1)(2j)^{2^k(2m+1)}}=-\sum_{j\geq 1}\sum_{h\geq 1}\frac{2}{2h(2j)^{2h}}=\sum_{j\geq 1}\log\left(1-\frac{1}{4j^2}\right)\tag{2}$$ and the claim follows from Wallis product.

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  • 2
    $\begingroup$ Nice catch! (the penultimate equality) (+1) $\endgroup$ – Olivier Oloa Mar 20 '17 at 14:23
  • 1
    $\begingroup$ Jack back at it again with the witchcraft $\endgroup$ – Gabriel Romon Mar 20 '17 at 14:29

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