Please help me to find a closed form for the infinite product $$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$ where $\tanh(z)=\frac{e^ze^{z}}{e^z+e^{z}}$ is the hyperbolic tangent.
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4Do you have any reason to believe that such a closed form exists? – George V. Williams May 13 '13 at 01:42

15http://www.wolframalpha.com/input/?i=product%28tanh%282%5En%29%5E%281%2F2%5En%29%2Cn%3D1..inf%29 vs. http://www.wolframalpha.com/input/?i=1exp%284%29 It could be a coincidence though... – Vladimir Reshetnikov May 13 '13 at 01:51

3Out of curiosity, where are you finding all of these questions? – Jemmy May 13 '13 at 02:32

1@Jeremy A friend of mine shared these problems with me. They were submitted to a math competition for students, but were rejected by the committee for various reasons: too hard, too easy, have been published before, not interesting etc. – Laila Podlesny May 13 '13 at 17:21

1@LailaPodlesny May I trouble you by requesting to know which math competition you are referring to? – Panglossian Oporopolist Sep 24 '15 at 12:03
2 Answers
For $x < 1$, we have the Taylor series expansion: $$f(x):= \frac{1}{4} \log \left( \frac{x  x^{1}}{x + x^{1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots $$
Then
$$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + \frac{x^{10}}{10} + \ldots $$ $$=  \frac{1}{2} \log(1  x^2).$$
Now let $x = e^{2}$. Then
$$ \log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) = \frac{1}{2^n} \log \left( \frac{e^{2^n}  e^{2^n}}{e^{2^n} + e^{2^n}}\right) $$ $$= \frac{4}{2^n} f(e^{2^n}) = \frac{4}{2^{n}} f(x^{2^{n1}}),$$
Hence summing over all $n \ge 1$, we see that, if the product is $P$, then
$$\log P = 4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n1}}) = 2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1  x^2),$$
and thus
$$P = \exp \log(1  x^2) = 1  x^2 = 1  e^{4}.$$
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2This is a very careful and crafty derivation. Excellent. I especially like where you pulled the ol' $\sum \log \leftrightarrow \log \prod$. – Coffee_Table May 13 '13 at 20:03
Let $$ f(x)=\prod_{n=0}^\infty\left(1x^{2^n}\right)^{1/2^n}\tag{1} $$ and $$ g(x)=\prod_{n=0}^\infty\left(1+x^{2^n}\right)^{1/2^n}\tag{2} $$ Then $$ \begin{align} f(x)\,g(x) &=\prod_{n=0}^\infty\left(1x^{2^{n+1}}\right)^{1/2^n}\\ &=\prod_{n=1}^\infty\left(1x^{2^n}\right)^{2/2^n}\\ &=\left(\frac{f(x)}{1x}\right)^2\tag{3} \end{align} $$ from which we get $$ \frac{f(x)}{g(x)}=(1x)^2\tag{4} $$ Note that $$ \prod_{n=1}^\infty\left(1x^{2^n}\right)^{1/2^n}=\frac{f(x)}{1x}\tag{5} $$ and $$ \prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}=\frac{g(x)}{1+x}\tag{6} $$ Therefore, combining $(4)$, $(5)$, and $(6)$, we get $$ \frac{\displaystyle\prod_{n=1}^\infty\left(1x^{2^n}\right)^{1/2^n}}{\displaystyle\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}}=1x^2\tag{7} $$ Plug $x=e^{2}$ into $(7)$ to get $$ \prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1e^{4}\tag{8} $$
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This looks much simpler than the accepted answer (which is also excellent) (+1). – Paramanand Singh Nov 19 '13 at 16:20

@ParamanandSingh: thanks. This question was just pointed out to me. However, new answers to old questions generally don't get as much attention as the older answers got. – robjohn Nov 19 '13 at 16:32