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I am currently working on a proof with a good friend of mine that involves adding more and more triangles to the sides of a regular polygon but keeping the longest diagonal constant until eventually, it becomes a circle. And we ended up with this formula.

$4$-sided regular$ to 8$-sided regular$to 16$-sided regular$to 32$-sided regularto ldots to n$-sided regular

(When $n$ tends to infinity, the area will be equal to that of a circle with the longest diagonal as diameter)

Moreover, we have used our calculator to input the numbers and we get the value of $3.140ldots$ which is very close to π But we can't be completely sure that the infinity sum really equals to π.

That is why we really need your knowledge of Maths to solve this.

Thanks in advance and Happy Holidays Everyone! :D

So the question is:
Is there a way to legitly prove that

$$sum_{n=0}^infty 2^n left(2cdot sinfrac{90^circ}{2^n}-sin frac{180^circ}{2^n}right)=pi$$

(please kindly refer to Appendix 1)

P.S. I apologize for my poor penmanship...

P.P.S. We tried using Wolfram Alpha, it didn't show us the step-by-step calculations

begin{align}
sum_{n=0}^m 2^n left(2 sin frac{90^circ}{2^n} - sin frac{180^circ}{2^n} right) &=sum_{n=0}^m 2^n left(2 sin frac{pi}{2^{n+1}} - sin frac{pi}{2^n} right)\
&=sum_{n=0}^m left(2^{n+1} sin frac{pi}{2^{n+1}} - 2^nsin frac{pi}{2^n} right)\
&=left(2^1 cdot sin frac{pi}{2} -2^0cdot sin frac{pi}{2^0}right)\
&+left(2^2 cdot sin frac{pi}{2^2} -2^1cdot sin frac{pi}{2^1}right)\
&+left(2^3 cdot sin frac{pi}{2^3} -2^2cdot sin frac{pi}{2^2}right)\
&vdots\
&+left(2^{m+1} sin frac{pi}{2^{m+1}} - 2^msin frac{pi}{2^m} right)\
&= 2^{m+1}sin frac{pi}{2^{m+1}}-2^0 sin pi \
&= 2^{m+1}sin frac{pi}{2^{m+1}}
end{align}

Hence

begin{align}sum_{n=0}^infty 2^n left(2 sin frac{90^circ}{2^n} - sin frac{180^circ}{2^n} right)&= lim_{m to infty}sum_{n=0}^m 2^n left(2 sin frac{90^circ}{2^n} - sin frac{180^circ}{2^n} right) \
&=lim_{m to infty} 2^{m+1}sin frac{pi}{2^{m+1}}\
&=lim_{m to infty} 2^{m+1}cdot frac{pi}{2^{m+1}}cdot frac{sin frac{pi}{2^{m+1}}}{frac{pi}{2^{m+1}}}\
&=lim_{m to infty} 2^{m+1}cdot frac{pi}{2^{m+1}}cdot lim_{m to infty}frac{sin frac{pi}{2^{m+1}}}{frac{pi}{2^{m+1}}}\
&= pi cdot 1\
&= piend{align}

Hint: $sin frac{pi}{2^n} = 2sin frac{pi}{2^{n+1}} cos frac{pi}{2^{n+1}}$ and the fundamental limit for sine tells us that $pi lim_{nto infty} frac{sin frac{pi}{2^{n+1}}}{frac{pi}{2^{n+1}}} = pi$. Try to do some algebraic manipulation to get those.

The sum telescopes, as the general term is (in radians)

$$2^{n+1}sinfracpi{2^{n+1}}-2^nsinfracpi{2^{n}}.$$

So the sum between $0$ and $m$ is

$$2^{m+1}sinfracpi{2^{m+1}}-sinpi=2^{m+1}left(fracpi{2^{m+1}}+oleft(frac1{2^{m+1}}right)right)$$

which converges to $pi$.