Solution to Six Years Old Olbers Paradox

So with the help of many sources I tried a solution.... To Question 🥁
Assume a static universe in which the stars are uniformly distributed. Divide all space into shells of constant thickness; the stars in any one shell by themselves subtend a solid angle of ω0
Allowing for the blocking out of distant stars by nearer stars, show that the total net solid angle subtended by all stars, shells extending to infinity, is exactly 4π
[ From 1.1 Infinite Series Mathematics Methods For Physicists Afrken and Weber]*

As it seems the problem is in geometric progression.

We apply the sum of geometric series formula.

With ω0 being initial parameter and r being shining factor which already seems to be less than 1, as question mentions; due to blocking ω goes on decreasing...

$$ {\lim_{{n}\rightarrow{\;∞\;}}}S=\frac{ω_{o}}{1−r} $$ I tried to find the exact factor but it seems very difficult visually to solve the equation even Chatgpt had hard time So I directly assumed... $$ {\lim_{{n}\rightarrow{\;∞\;}}}S_{n}=4π $$ Thus now if we solve for r $$ 4π=\frac{ω_{o}}{1−r} $$ $$ r=1−\frac{ω_{o}}{4π} $$ Feel free advice any changes

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