# Infinite Series that $= \pi$ or $= \pi^2$

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

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

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

It was derived from:

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

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

# Algebra

Note by A Former Brilliant Member
10 months, 1 week ago

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

- 9 months, 4 weeks ago

Ok... @Aruna Yumlembam

- 9 months, 4 weeks ago

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

- 9 months, 4 weeks ago

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

- 9 months, 4 weeks ago

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.

- 9 months, 4 weeks ago