I see this formula given below on You tube video of mathologer channel and then I try to find some new method to prove it:
$$\sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$$
I tried to prove it geometrically like this
Our attempt:
(1) First I tried to convert it in inverse trignometric form like this but that doesn't help much:
(2) In my second attempt I rotate the length of $1/2$ length from $1$ then I rotate $1/3$ length from remaining $1/2$ but that thing doesn't help us.
(3) In my third attempt, I tried to use coordinate geometry but that makes things more complex.
My question:How to prove that that summation of $1/n^2$ where $n$ tends to infinity is equal to $π^2/6$ by using spiral right angle triangle method?
EDIT
NOTE: Sinc the last line segment Whose length tends to Square root of $π^2/6$ but not exactly equal to Square root of $π^2/6$ so it is probably not possible to solved it by using pure geometry. we understand that there must be needs of theory of Limit to prove it . so we will also accept the solution which take the use of both concept means geometry with slight use of calculus.
Since, using bgeneralized harmonic numbers $$\sum_{i=1}^k \frac 1 {i^2}=H_k^{(2)}$$
quoting @Jean Marie "your spiral will turn an infinite number of turns..." $$A_k=\tan ^{-1}\left(\frac{1}{(k+1) \sqrt{H_k^{(2)}}}\right)$$ which, for large $k$ is $$A_k=\frac{\sqrt{6}}{\pi k}\Bigg[1-\frac{\pi ^2-3}{\pi ^2 k}+\frac{27-13 \pi ^2+2 \pi ^4}{2 \pi ^4 k^2}-\frac{-135+90 \pi ^2-22 \pi ^4+2 \pi ^6}{2 \pi ^6 k^3}+O\left(\frac{1}{k^4}\right) \Bigg]$$
Repeating @Ivan Kaznacheyeu's calculations $$\sum_{k=1}^{6400} A_k=6.678$$ while using the truncated series, we obtain $6.607$
In fact, computed exactly $$\sum_{k=1}^{3857} A_k=2\pi +0.0000601086$$