Infinite Sum of an Inverse Trig Expression

Question

I am attempting to find either a closed form for the following infinite sum, or failing that, the value $p$ for which the sum converges to $2\pi$ (somewhere around $0.82$?). $$\sum_{i=1}^\infty \arccos \left( \frac{1 + p^i+p^{i+1}-p^{2i+1}}{1 + p^i+p^{i+1}+p^{2i+1}} \right) $$

Thank you ahead of time for your help.

Later edit: I put together a post with further context and thoughts about the problem I was trying to solve for those interested. The "object" I was asking about looks like this: Please let me know if this is already well-trod so I can read more + reference those who have already done this work.

Answer 1

If you are patient, work using only rational numbers and arbitrary precision.

Perform a binary search for solving, as close as possible the equation $$f(p)=-2\pi+\sum_{i=1}^\infty \arccos \left(1-\frac{2 p^{2 i+1}}{\left(1+p^i\right) \left(1+p^{i+1}\right)} \right)=0$$

First, notice that, if $p <1$ the series is convergent since $$\arccos \left(1-\epsilon \right)=\sqrt{2 \epsilon }+O\left(\epsilon ^{3/2}\right)$$

Then $$a_i=\arccos \left(1-\frac{2 p^{2 i+1}}{\left(1+p^i\right) \left(1+p^{i+1}\right)} \right)\quad \implies \quad \lim_{i\to \infty } \, \frac {a_{i+1}}{a_{i}}=p$$

For $p=0.82$ we have (with $R^2=0.99977$) $$\log(a_i) \sim 0.463189- 0.196468\, i$$

All of the above let thinking about a quite fast convergence.

You could find $$f\left(\frac{823930515\color{red}{69999}}{100000000000000}\right)=-1.042\times 10^{-9}$$ $$f\left(\frac{823930515\color{red}{70000}}{100000000000000}\right)=+8.229\times 10^{-10}$$ This number is not recognized by inverse symbolic calculators.

Edit

Newton method seems to be quite efficient for this problem. Using $p_0=\frac {41}{50}$, the successive iterates are $$\left( \begin{array}{cc} n & p_n \\ 0 & 0.82000000 \\ 1 & 0.82401860 \\ 2 & 0.82393056 \\ 3 & 0.82393052 \end{array} \right)$$