MathWorld states that
$\displaystyle \lim_{j\to\infty}\prod_{n=j}^{2j}\frac{\pi}{2\tan^{-1}n}=4^{\frac{1}{\pi}}$ (equation 130)
With a calculator, I find a general formula:
$\displaystyle \lim_{j\to\infty}\prod_{n=j}^{mj}\frac{\pi}{2\tan^{-1}(kn)}=m^{\frac{2}{k\pi}}$
But I have not found a proof for this. Any proof would be appreciated.
It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write
$$ P_j = P_j(m,k) = \prod_{n=j}^{mj} \frac{\pi}{2\arctan(kn)} $$
for the expression inside the limit. Then using the relation $\arctan(x) = \frac{\pi}{2} - \arctan(\frac{1}{x})$ which holds for $x > 0$, we find that
$$ \log\left(\frac{\pi}{2\arctan(x)}\right) = -\log\left(1 - \tfrac{2}{\pi}\arctan\left(\tfrac{1}{x}\right) \right) = \tfrac{2}{\pi x} + \mathcal{O}\left(x^{-2}\right) \quad \text{as} \quad x \to \infty.$$
So it follows that
$$ \log P_j = \sum_{n=j}^{mj} \left( \log\frac{\pi}{2} - \log\arctan(kn)\right) = \sum_{n=j}^{mj} \left( \frac{2}{\pi kn} + \mathcal{O}(n^{-2}) \right). $$
By noting that this is a Riemann sum with a vanishing error term, we realize that $\log P_n$ converges to $\int_{1}^{m} \frac{2}{\pi kx} \, dx = \frac{2}{\pi k}\log m$ as $n\to\infty$. Exponentiation then yields $ P_n \to m^{\frac{2}{\pi k}} $ as expected.