Show that $\lim_{n \to \infty}\prod_{k = n}^{2n}\dfrac{\pi}{2\tan^{-1}k} = 4^{1/\pi}$

Question

MathWorld states that (see equation $(130)$) $$\lim_{n \to \infty}\prod_{k = n}^{2n}\dfrac{\pi}{2\tan^{-1}k} = 4^{1/\pi}$$ and attributes it to Gosper. I believe an approach to establish the formula is to take logs to get $$\lim_{n \to \infty}\sum_{k = n}^{2n}\log\left(\frac{\pi}{2\tan^{-1}k}\right) = \frac{\log 4}{\pi}$$ But I have not been able to express the LHS as as integral. Perhaps some manipulation of the LHS is needed. Any proof or hints would be helpful.

Answer 1

HINT: $$ \lim_{n \to \infty}\sum_{k = n}^{2n}\log\left(\frac{\pi}{2\tan^{-1}k}\right)= \lim_{n \to \infty}\sum_{k = n}^{2n}\log\left(\frac{\pi}{2\left(\frac\pi2-\tan^{-1}\frac1k\right)}\right)=\lim_{n \to \infty}\sum_{k = n}^{2n}\log\left(1+\frac{2}{\pi k}\right)$$ due to $$ \tan^{-1}\frac1k=\frac1k+O(\frac{1}{k^3})$$ as $k$ is great.

Edit: Since, for $\alpha>1$, each series of the form $$\sum_{k =1}^{\infty}\frac{1}{k^{\alpha}} $$ is convergent, then we have $$\lim_{n \to \infty}\sum_{k = n}^{2n}O(\frac{1}{k^{\alpha}})=0$$ and we are allowed to write the second equality above.