TITLE: Show that $\lim_{n\to \infty}\prod_{i=n}^{Bn}\frac{\arctan(i\phi)}{\arccos\left(\frac{\phi}{i}\right)}=B^{\frac{2}{\pi}}$
QUESTION [5 upvotes]: Inspired from Gosper's formula
$$\lim_{n\to \infty}\prod_{i=n}^{2n}\frac{\pi}{2\arctan(i)}=4^{\frac{1}{\pi}}$$
(See Pi Formulas on MathWorld)
Through mathematical experimental we found another formula similar to that of Gosper
$\phi=\frac{1+\sqrt5}{2}$
B is an integer, $B\ge2$
$$\lim_{n\to \infty}\prod_{i=n}^{Bn}\frac{\arctan(i\phi)}{\arccos\left(\frac{\phi}{i}\right)}=B^{\frac{2}{\pi}}$$
We need help on proving it. Can anyone help us to prove it.

REPLY [7 votes]: Denote $f(x)=\log\frac{\arctan(x\phi)}{\arccos\left(\frac{\phi}{x}\right)}$. Expanding into series
$$
f(x)=\frac{2}{\pi  x}+O(x^{-2}),\quad x\to\infty,
$$
and using the Euler's approximation for the harmonic sums
$$
\sum_{i=1}^n\frac1i=\gamma+\log n+O(1/n),
$$
where $\gamma$ is the Euler's constant, gives
$$
\log\lim_{n\to \infty}\prod_{i=n}^{Bn}\frac{\arctan(i\phi)}{\arccos\left(\frac{\phi}{i}\right)}=
\lim_{n\to \infty}\sum_{i=n}^{Bn}f(i)=
\lim_{n\to \infty}\left[\frac{2}{\pi}\left(\sum_{i=1}^{Bn}\frac1i-\sum_{i=1}^{n-1}\frac1i\right)+O(n^{-1})\right]=
$$
$$
=\lim_{n\to \infty}\left[\frac{2}{\pi}(\log Bn-\log n)+O(n^{-1})\right]=\frac{2}{\pi}\log B.
$$