Why Doesn't this equation work?

Question

$$\frac{n2^nn!^2}{(2n)!}$$

Is supposed to output the n^th decimal of pi, this works fine with $$n = 1$$, but why not with $$n = 2$$? (I ommited the sigma, equation was found here)

Answer 1

That equation is not any digit extraction formula for π, it is the expression for π that Plouffe began with to develop his decimal digit extraction algorithm. For an equation to correspond directly to a digit extraction algorithm it must be a BBP-type forumla, which have this form: $$\alpha = \sum_{k=1}^{\infty}{\frac{f(k)}{b^k}}$$

The original BBP formula is a digit extraction formula for π, but with $b=16$, so it extracts hexadecimal digits, not decimal digits. Plouffe (the 'P' in BBP) did publish a decimal digit extraction algorithm, but it does not correspond to a BBP-type formula.

Plouffe's paper (2009 reprint) is focused on the underlying principles, and does not give a straightforward description of the algorithm. The 2009 reprint references two improved decimal digit extraction algorithms, which have somewhat clearer descriptions:

• Bellard's refinement of Plouffe's algorithm, with a somewhat more straightforward description of the refined algorithm
• Gourdon's completely different algorithm has a section dedicated to describing the completely different algorithm

Answer 2

A known result related to $\pi$ and your expression is : $$\sum_{n=1}^\infty\frac{n\,2^n\,n!^2}{(2n)!}=\pi+3$$

Gery Huvent conjectured (equation $(401)$ from this excellent $\pi$ site) that $$\forall k\ge 0,\quad\sum_{n=1}^\infty\frac{n^k\,2^n\,n!^2}{(2n)!}=a_k\,\pi+b_k\quad\text{with}\;\frac {b_k}{a_k}\to\pi$$

The fractions $\dfrac {b_k}{a_k}$ obtained for the first values of $k$ are : \begin{array} {l|cc} k&&\frac {b_k}{a_k}\\ \hline 1&\frac 31&=&3\\ 2&\frac {11}{7/2}&=&\frac {22}7\\ 3&\frac{55}{35/2}&=&\frac {22}7\\ 4&\frac{355}{113}&=&\frac{355}{113}\\ \end{array}