Infinite Series that =π= \pi or =π2= \pi^2

Yesterday night, I was looking for an infinite series that equals to π\pi. And I found one! Using WolframAlpha's Infinite Series Calculator, I got: It was derived from: n=01(4n+1)(4n+3)=π8\displaystyle\sum_{n=0}^{\infty} \frac{1}{(4n + 1)(4n + 3)} = \frac{\pi}{8}

I also found an infinite series that converges to π2\pi^2:

n=06(n+1)2=π2\displaystyle\sum_{n=0}^{\infty} \frac{6}{(n + 1)^2} = \pi^2

It was derived from:

n=01(n+1)2=π26\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n + 1)^2} = \frac{\pi^2}{6}

I hope you enjoyed this article and comment if you find any other series that converges to π2\pi^2 or π\pi - must be in LaTeX!



Note by A Former Brilliant Member
7 months, 2 weeks ago

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π\pi=n=1Γ(n1/2)Γ(1/2+1)Γ(n+1)\sum_{n=1}^{\infty}\frac{\Gamma(n-1/2)\Gamma(1/2+1)}{\Gamma(n+1)}.Hope you like it.

Aruna Yumlembam - 7 months, 1 week ago

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Ok... @Aruna Yumlembam

A Former Brilliant Member - 7 months, 1 week ago

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The proof relies on the iterated properties of beta function.Check the discussion.

Aruna Yumlembam - 7 months ago

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@Aruna Yumlembam I did - it's great but I am in Year 10, becoming Year 11. How can I do calculus then?

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Well I didn't discovered calculus until I was at last year of becoming 10.You know it's never too late to learn something new.

Aruna Yumlembam - 7 months ago

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