Come On Bois {0}

First, note that for $n>0$ and $2h := \pi - x$ , $4\cos^2\frac{x}{2} = 4\cos^2\left(\frac{\pi}{2}-h\right) = 4\sin^2h$ $\begin{aligned}\frac{n\cosh(nx)}{\sinh(n\pi)} &= n\cdot \frac{e^{nx} + e^{-nx}}{e^{n\pi}-e^{-n\pi}} \\ &= n\cdot \frac{e^{n(\pi-x)-2n\pi} + e^{-n(\pi-x)}}{1-e^{-2n\pi}} \\ &= \sum_{k=0}^\infty n\left(e^{2nh-2n\pi}+e^{-2nh}\right)\cdot e^{-2nk\pi} \\ &= \sum_{k=0}^\infty n\left(e^{2h-2\pi(k+1)}\right)^n + n\left(e^{-2h-2k\pi}\right)^n\end{aligned}$ Therefore, since all the terms are positive, we can interchange the order of summation to write $\begin{aligned} \sum_{n=1}^\infty \frac{n\cosh(nx)}{\sinh(n\pi)} &= \sum_{k=0}^\infty \sum_{n=1}^\infty n\left(e^{2h-2\pi(k+1)}\right)^n + n\left(e^{-2h-2k\pi}\right)^n & \\ &= \sum_{k=0}^\infty \frac{e^{2h-2\pi(k+1)}}{\left(1-e^{2h-2\pi(k+1)}\right)^2} + \frac{e^{-2h-2k\pi}}{\left(1-e^{-2h-2k\pi}\right)^2}& \text{for } 0 < h < \pi \\ &= \frac{e^{2h}}{(1 - e^{2h})^2} + \sum_{k=1}^\infty \frac{e^{2h-2k\pi}}{\left(1-e^{2h-2k\pi}\right)^2} + \frac{e^{-2h-2k\pi}}{\left(1-e^{-2h-2k\pi}\right)^2} \\ &= \frac{1}{(e^{-h}-e^{h})^2} + \sum_{k=1}^\infty \frac{1}{\left(e^{-(h-k\pi)}-e^{h-k\pi}\right)^2} + \frac{1}{\left(e^{h+k\pi}-e^{-(h+k\pi)}\right)^2} \\ &= \frac{1}{4\sinh^2h} + \frac{1}{4} \sum_{k=1}^\infty \frac{1}{\sinh^2(h-k\pi)} + \frac{1}{\sinh^2(h+k\pi)} \end{aligned}$

Now, I can finally consider the limit. In terms of $h$ and using our work above, it becomes $\frac{1}{4}\lim_{h\to 0^+}\left(\frac{1}{\sin^2h} - \frac{1}{\sinh^2h} - \sum_{k=1}^\infty \frac{1}{\sinh^2(h-k\pi)} - \sum_{k=1}^\infty \frac{1}{\sinh^2(h+k\pi)}\right)$ Looking at the limit with just the first two terms: $\begin{aligned} \lim_{h\to 0^+} \frac{1}{\sin^2h} - \frac{1}{\sinh^2h} &= \lim_{h\to 0^+} \frac{\sinh^2h - \sin^2h}{\left(\sin^2h\right)\left(\sinh^2h\right)} \\ &= \lim_{h\to 0^+} \frac{\left(h+\frac{1}{3!}h^3 + o(h^3)\right)^2 - \left(h - \frac{1}{3!}h^3 + o(h^3)\right)^2}{\left(h+\frac{1}{3!}h^3 + o(h^3)\right)^2\left(h - \frac{1}{3!}h^3 + o(h^3)\right)^2} \\ &= \lim_{h\to 0^+} \frac{\left(h^2 + \frac{1}{3}h^4 + o(h^4)\right)-\left(h^2 - \frac{1}{3}h^4 + o(h^4)\right)}{\left(h^2 + \frac{1}{3}h^4 + o(h^4)\right)\left(h^2 - \frac{1}{3}h^4 + o(h^4)\right)} \\ &= \lim_{h\to 0^+} \frac{\frac{2}{3}h^4 + o(h^4)}{h^4 + o(h^4)} \\ &= \frac{2}{3} \end{aligned}$ Also, using the monotone convergence theorem (note that $\sinh^2(h-k\pi) = \sinh^2(k\pi-h) > \sinh^2((k-1)\pi)$ ), $\lim_{h\to 0^+} \sum_{k=1}^\infty \frac{1}{\sinh^2(h-k\pi)} = \lim_{h\to 0^+} \sum_{k=1}^\infty \frac{1}{\sinh^2(h+k\pi)} = \sum_{k=1}^\infty \frac{1}{\sinh^2(k\pi)}$

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As a summary so far, we have shown that the limit in the question equals $\frac{1}{4}\left(\frac{2}{3} - 2\sum_{k=1}^\infty \frac{1}{\sinh^2(k\pi)}\right) = \frac{1}{6} - \frac{1}{2}\sum_{k=1}^\infty \frac{1}{\sinh^2(k\pi)}$ and here, I have to be honest, I don't know how to find the exact sum of that last series, but I do know that it converges quickly as for all $k\geq 1$ , $\sinh^2(k\pi) = \frac{1}{2}\left(\cosh(2k\pi) - 1\right) > \frac{1}{4}e^{2k\pi} - \frac{1}{8}e^{2k\pi} = \frac{1}{8}e^{2k\pi} > 0,$ so that $\left|\frac{1}{2}\sum_{k=N+1}^\infty \frac{1}{\sinh^2(k\pi)}\right| < \sum_{k=N+1}^\infty 4e^{-2k\pi} = \frac{4e^{-2(N+1)\pi}}{1-e^{-2\pi}} = \frac{4e^{-2N\pi}}{e^{2\pi}-1}$

Therefore the error in $\frac{1}{6} - \frac{1}{2}\left(\frac{1}{\sinh^2\pi} + \frac{1}{\sinh^2(2\pi)}\right) \approx 0.162910818$ is at most $\left(4e^{-4\pi}\right)/\left(e^{2\pi}-1\right) < 2.7 \times 10^{-8}$ so we know the answer is at least accurate when rounded to $\boxed{0.1629108}$