Sum of Infinity of Trigo to Pi

Question

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$-sided regular\to \ldots \to n$-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 $3.140\ldots$ 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

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

(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

Answer 1

\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^n\sin \frac{\pi}{2^n} \right)\\ &=\left(2^1 \cdot \sin \frac{\pi}{2} -2^0\cdot \sin \frac{\pi}{2^0}\right)\\ &+\left(2^2 \cdot \sin \frac{\pi}{2^2} -2^1\cdot \sin \frac{\pi}{2^1}\right)\\ &+\left(2^3 \cdot \sin \frac{\pi}{2^3} -2^2\cdot \sin \frac{\pi}{2^2}\right)\\ &\vdots\\ &+\left(2^{m+1} \sin \frac{\pi}{2^{m+1}} - 2^m\sin \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\\ &= \pi\end{align}

Answer 2

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

Answer 3

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

$$2^{n+1}\sin\frac\pi{2^{n+1}}-2^n\sin\frac\pi{2^{n}}.$$

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

$$2^{m+1}\sin\frac\pi{2^{m+1}}-\sin\pi=2^{m+1}\left(\frac\pi{2^{m+1}}+o\left(\frac1{2^{m+1}}\right)\right)$$

which converges to $\pi$.