I discovered a sequence that should converge to $\pi$ , but how to prove that it really converges to $\pi$?

Asked May 03 '16 at 14:19

Active May 03 '16 at 14:19

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So yesterday I took paper and pencil and did some constructions of geometrical nature and I discovered a sequence that should converge to $\pi$.

It goes like this:

Define $a_1=\sqrt{2}$ and for every other $n \in \mathbb N$ define $a_n=\sqrt{2+a_{n-1}}$. Define $S_n=2^n \cdot \sqrt{2-a_{n-1}}$. Then we should have $\lim_{n \to \infty}S_n=\pi$.

This formula (more correctly, sequence) can be seen on this page and is labeled with number (66) so I did not discover anything new.

I found this sequence but I did not rigorously prove that it converges to $\pi$, the story is that I saw a pattern for the first few natural numbers and then defined the expression for general $n$ and when I did that I found the sequence on the above linked page.

My question would be:

In which way would you prove that this sequence converges to $\pi$?

asked May 03 '16 at 14:19

Farewell

5,046

1

A trigonometric proof is the best way, since this is related to the famous Vieta formula (look it up). Use the doulbe angle formula for the sine and cosine functions – Yuriy S May 03 '16 at 14:22
2

See this question for the exact same limit. There is a lot of answers there – Yuriy S May 03 '16 at 14:24
@YuriyS Thank you, that looks helpful. – Farewell May 03 '16 at 14:25

I discovered a sequence that should converge to $\pi$ , but how to prove that it really converges to $\pi$?

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