Suppose I have \(\angle ABC\), and I draw out an angle bisector of \(\angle ABC\), such that \(\angle ABA_{1} = \angle CBA_{1}\) as described above.

Next, I make an angle bisector of \(\angle CBA_{1}\), such that \(\angle A_{1}BA_{2} = \angle A_{2} BC\)

Next, I make an angle bisector of \( \angle A_{1}BA_{2}\), such that \(\angle A_{1}BA_{3} = \angle A_{3}BA_{2}\)

Next, I make an angle bisector of \(\angle A_{3}BA_{2}\), such that \(\angle A_{3}BA_{4} = \angle A_{4}BA_{2}\)

If I am to repeat this algorithm forever, find \[\lim _{ n\rightarrow \infty }{ \frac { \angle { A }_{ n }BC }{ \angle { A }BC } } \]

Give your answer to 3 decimal places.

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