# Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$

Please help me to find a closed form for the infinite product $$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$ where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$ is the hyperbolic tangent.

• Do you have any reason to believe that such a closed form exists? – George V. Williams May 13 '13 at 1:42
• wolframalpha.com/input/… vs. wolframalpha.com/input/?i=1-exp%28-4%29 It could be a coincidence though... – Vladimir Reshetnikov May 13 '13 at 1:51
• Out of curiosity, where are you finding all of these questions? – Jemmy May 13 '13 at 2:32
• @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
• @LailaPodlesny May I trouble you by requesting to know which math competition you are referring to? – Kugelblitz Sep 24 '15 at 12:03

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

$$\log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) = \frac{1}{2^n} \log \left( \frac{e^{2^n} - e^{-2^n}}{e^{2^n} + e^{-2^n}}\right)$$ $$= \frac{-4}{2^n} f(e^{-2^n}) = \frac{-4}{2^{n}} f(x^{2^{n-1}}),$$

Hence summing over all $n \ge 1$, we see that, if the product is $P$, then

$$\log P = -4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n-1}}) = -2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1 - x^2),$$

and thus

$$P = \exp \log(1 - x^2) = 1 - x^2 = 1 - e^{-4}.$$

• Nicely done! :) – Caran-d'Ache May 13 '13 at 4:47
• This 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 $$f(x)=\prod_{n=0}^\infty\left(1-x^{2^n}\right)^{1/2^n}\tag{1}$$ 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(1-x^{2^{n+1}}\right)^{1/2^n}\\ &=\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{2/2^n}\\ &=\left(\frac{f(x)}{1-x}\right)^2\tag{3} \end{align} from which we get $$\frac{f(x)}{g(x)}=(1-x)^2\tag{4}$$ Note that $$\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}=\frac{f(x)}{1-x}\tag{5}$$ 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 $$\frac{\displaystyle\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}}{\displaystyle\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}}=1-x^2\tag{7}$$ Plug $x=e^{-2}$ into $(7)$ to get $$\prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1-e^{-4}\tag{8}$$

• 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