Ramanujan Formula for Pi, WZ method

\displaystyle \frac{2}{\pi}=\sum_{k=0}^{\infty}(-1)^{k}(4 k+1) \frac{(1 / 2)_{k}^{3}}{k !^{3}}

This can be seen as specialization of the following identity at {n = -\frac{1}{2}.}

\displaystyle \frac{\Gamma(3 / 2+n)}{\Gamma(3 / 2) \Gamma(n+1)}=\sum_{k=0}^{\infty}(-1)^{k}(4 k+1) \frac{(1 / 2)_{k}^{2}(-n)_{k}}{k !^{2}(3 / 2+n)_{k}}

Proof: We provide a proof by WZ method.

\displaystyle F(n,k) =\frac{(-1)^{k}(4 k+1) \frac{(1 / 2)_{k}^{3}}{k !^{3}}}{\frac{\Gamma(3 / 2+n)}{\Gamma(3 / 2) \Gamma(n+1)}}

We want to find {G(n,k)} such that

\displaystyle F(n+1,k)-F(n, k) = G(n,k) -G(n, k-1)


If we have such a {G(n,k)}, then we have

\displaystyle \sum_{k} \left(F(n+1,k)-F(n, k) \right) = \sum_{k }\left(G(n,k) -G(n, k-1)\right)

and we can see that {\sum_{k} F(n,k)} will be a constant.

Choice found by algorithms:

\displaystyle G(n, k):=\frac{(2 k+1)^{2}}{(2 n+2 k+3)(4 k+1)} F(n, k)

Now

\displaystyle \sum_{k} F(n,k) = \sum_{k} F(0,k) =1

So we are done.

https://arxiv.org/pdf/math/9306213.pdf

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