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I found an interesting series
$$\sum_{k=1}^\infty \log \left(\tanh \frac{\pi k}{2} \right)=\log(\vartheta_4(e^{-\pi}))=\log \left(\frac{\pi^{\frac{1}{4}}}{2^{\frac{1}{4}}\Gamma \left( \frac{3}{4}\right)} \right)$$
Thank you very much!
We have $$\vartheta_{4}(q) = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2}\tag{1}$$ It is easily seen that the above product can be written as $$\prod_{n = 1}^{\infty}(1 - q^{2n})\cdot\frac{(1 - q^{n})^{2}}{(1 - q^{2n})^{2}} = \prod_{n = 1}^{\infty}\frac{(1 - q^{n})^{2}}{1 - q^{2n}} = \prod_{n = 1}^{\infty}\frac{1 - q^{n}}{1 + q^{n}}\tag{2}$$ Putting $q = e^{-\pi}$ and noting that $$\tanh(n\pi/2) = \frac{e^{n\pi/2} - e^{n\pi/2}}{e^{n\pi/2} + e^{n\pi/2}} = \frac{1 - e^{-n\pi}}{1 + e^{-n\pi}} = \frac{1 - q^{n}}{1 + q^{n}}\tag{3}$$ we have via virtue of equations $(1), (2), (3)$ $$\vartheta_{4}(e^{-\pi}) = \prod_{n = 1}^{\infty}\tanh\left(\frac{n\pi}{2}\right)$$ so that the first equality is proved.
Now to calculate the value of $\vartheta_{4}(q)$ for $q = e^{-\pi}$ note that $$\vartheta_{4}(q) = \frac{\vartheta_{4}(q)}{\vartheta_{3}(q)}\cdot \vartheta_{3}(q) = \sqrt{k'}\sqrt{\frac{2K}{\pi}}\tag{4}$$ For $q = e^{-\pi}$ we have $k = k' = 1/\sqrt{2}$ and $$K = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - (1/2)\sin^{2}x}} = \frac{1}{4\sqrt{\pi}}\Gamma^{2}\left(\frac{1}{4}\right)\tag{5}$$ and hence $$\vartheta_{4}(e^{-\pi}) = 2^{-3/4}\pi^{-3/4}\Gamma(1/4)$$ and noting that $$\Gamma(1/4)\Gamma(3/4) = \frac{\pi}{\sin(\pi/4)} = \sqrt{2}\pi$$ we get $$\vartheta_{4}(e^{-\pi}) = \frac{\pi^{1/4}}{2^{1/4}\Gamma(3/4)}$$ so that the second equality of the question is also proved.
The integral in $(5)$ is easily evaluated via the use of beta and gamma functions. Thus \begin{align} K &= K(1/\sqrt{2}) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - (1/2)\sin^{2}x}}\notag\\ &= \sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{2 - \sin^{2}x}}\notag\\ &= \sqrt{2}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1 + \cos^{2}x}}\notag\\ &= \sqrt{2}\int_{0}^{1}\frac{dt}{\sqrt{1 - t^{4}}}\text{ (by putting }t = \cos x)\notag\\ &= \frac{\sqrt{2}}{4}\int_{0}^{1}x^{-3/4}(1 - x)^{-1/2}\,dx\text{ (by putting }t^{4} = x)\notag\\ &= \frac{\sqrt{2}}{4}B\left(\frac{1}{4}, \frac{1}{2}\right)\notag\\ &= \frac{\sqrt{2}}{4}\cdot\dfrac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)}\notag\\ &= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)}\notag\\ &= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma^{2}\left(\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{3}{4}\right)}\notag\\ &= \frac{\sqrt{2\pi}}{4}\cdot\dfrac{\Gamma^{2}\left(\frac{1}{4}\right)}{\dfrac{\pi}{\sin(\pi/4)}}\notag\\ &= \frac{1}{4\sqrt{\pi}}\Gamma^{2}\left(\frac{1}{4}\right)\notag \end{align}
Update: Apparently I forgot to shed some light on the second question asked by OP namely any references/books on the theory of Jacobi Elliptic and Theta functions. This I deal with now.
The theory of Elliptic functions and theta functions is a very fascinating one. My own sources of study in this field are the following books (order of the books listed is not important here):
All the above books combined together cover most of the elementary and some advanced topics related to elliptic and theta functions except its link with the algebraic number theory (mainly the link with imaginary quadratic extensions of $\mathbb{Q}$). This is perhaps the most important and deep topic which is now famous by the name of modular forms. Unfortunately I don't have much knowledge on this topic.
The references listed above can be found online for free (if you search enough). I have tried to extract material from these references and present a coherent theory of elliptic integrals/functions and theta functions in my blog posts (in fact I started blogging only to document whatever I had learnt about these mysterious theta functions). The blog also contains the Ramanujan's theory of theta functions and the Borwein's approach via Arithmetic Geometric Mean. The advantage of the blog posts is that they are concise and are written in a particular order such that pre-requisites for proving a result are discussed before presenting the result.
It may be of interest to note that the sum $$ g(x) = \sum_{k=1}^\infty \log\tanh (kx)$$ is harmonic and may be evaluated using Mellin transforms, yielding an asymptotic expansion about zero.
The Mellin transform $f^*(s)$ of the base function $$ f(x) = \log\tanh x$$ may be computed as follows
\begin{align} f^*(s) & = \mathfrak{M}(f(x); s) = \int_0^\infty \log \frac{e^x-e^{-x}}{e^x+e^{-x}} x^{s-1} \, dx = \int_0^\infty \log \left(1 - 2\frac{e^{-x}}{e^x+e^{-x}}\right) x^{s-1} dx \\[6pt] & = \int_0^\infty \log \left(1 - 2\frac{e^{-2x}}{1+e^{-2x}}\right) x^{s-1} dx = - \int_0^\infty \sum_{q\ge 1} \frac{1}{q} 2^q e^{-2qx}\left(\frac{1}{1+e^{-2x}}\right)^q x^{s-1} dx \\[6pt] & = -\sum_{q\ge 1} \frac{1}{q} 2^q \int_0^\infty e^{-2qx} \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} e^{-2mx} x^{s-1} dx \\[6pt] & = -\sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \int_0^\infty e^{-2qx} e^{-2mx} x^{s-1} dx \\[6pt] & =-\Gamma(s)\sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \frac{1}{(2m+2q)^s} \\[6pt] & = -\frac{\Gamma(s)}{2^s} \sum_{q\ge 1} \frac{1}{q} 2^q \sum_{m\ge 0} (-1)^m \binom{m+q-1}{m} \frac{1}{(m+q)^s}. \end{align}
To complete this calculation, ask about the coefficient of $$\frac{1}{n^s} = \frac{1}{(m+q)^s}.$$ It is given by $$\sum_{m=0}^{n-1} \frac{1}{n-m} 2^{n-m} (-1)^m \binom{n-1}{m} = - \frac{1}{n} (-1)^n + \frac{1}{n} \sum_{m=0}^n 2^{n-m} (-1)^m \binom{n}{m} = \frac{1}{n} \left(1-(-1)^n\right).$$ It follows that the double sum is $$\sum_{n\ge 1} \frac{1-(-1)^n}{n^{s+1}}= 2 \sum_{m\ge 0} \frac{1}{(2m+1)^{s+1}} = 2 \zeta(s+1) \left(1 - \frac{1}{2^{s+1}}\right) = \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$ This gives the following for $f^*(s):$ $$ f^*(s) = -\frac{\Gamma(s)}{2^s} \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$ Now introduce $g(x)$, the harmonic sum we are trying to calculate, so that $$ g(x) = \sum_{k=1}^\infty f(kx).$$ The Mellin transform of $g(x)$ is then given by $$ g^*(s) = \mathfrak{M}(g(x); s) = -\frac{\Gamma(s)}{2^s} \zeta(s) \zeta(s+1) \left(2 - \frac{1}{2^s}\right).$$ The zeros of the two zeta function terms cancel the poles of the gamma function, so that inverting $g^*(s)$ we only have two terms that contribute, namely $$\operatorname{Res}(g^*(s) x^{-s}; s=1) = -1/8\,{\frac {{\pi }^{2}}{x}}$$ and $$\operatorname{Res}(g^*(s) x^{-s}; s=0) = 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \left( x \right).$$ This yields that in a neighborhood of zero $$ g(x) \sim 1/2\,\log \left( 2\,\pi \right) -1/2\,\log \left( x \right) -1/8\,{\frac {{\pi }^{2}}{x}}$$ Setting $x=\frac{\pi}{2}$, we obtain that $$g\left(\frac{\pi}{2}\right) \sim \log 2 - \frac{\pi}{4},$$ which produces only three good digits. On the other hand, for e.g. $x=\frac{1}{4}$, we get $$g\left(\frac{1}{4}\right) \sim 3/2\,\log \left( 2 \right) +1/2\,\log \left( \pi \right) -1/2\,{\pi }^{2} \sim -3.322716487,$$ which has nine good digits.
For e.g. $x=\frac{\pi}{16}$, we get $$g\left(\frac{\pi}{16}\right) \sim 5/2\,\log \left( 2 \right) -2\,\pi \sim -4.5503173557797232034$$ which has 20 good digits.
It seems quite intriguing to ask whether this expansion can also be derived directly from properties of the Jacobi theta function without using Mellin transforms.
