Pi Approximation Day

[Albert Chan]

(07-23-2022 02:20 AM)Thomas Klemm Wrote:  It uses Ramanujan's formula:

\(
\begin{align}
\pi^2 = 10 - \sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}
\end{align}
\)

We can adjust numerator, to speed up partial fraction decomposition.

1 = ((n+1)-n)^2 = (n+1)^2 - 2*n*(n+1) + n^2
1 = ((n+1)-n)^3 = (n+1)^3 - 3*n*(n+1) - n^3

1 = (n+1)^3 - n^3 - 3*n*(n+1) * ((n+1)^2 - 2*n*(n+1) + n^2)

Divide both side by (n*(n+1))^3, we have:

\(\displaystyle \frac{1}{n^3\;(n+1)^3} =
\left(\frac{1}{n^3}-\frac{1}{(n+1)^3} \right)
- 3×\left( \frac{1}{n^2} + \frac{1}{(n+1)^2}\right)
+ 6×\left( \frac{1}{n} - \frac{1}{n+1}\right)
\)

\(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^3(n+1)^3}
= 1 - 3×(\frac{\pi^2}{6} + \frac{\pi^2}{6} - 1) + 6 = 10 - \pi^2\)