# Accelerated Wallis' product

Recently I have obtained the following:

$$\prod_{k=1}^{\infty} \left(1 + \frac 1{4\cdot k^2\cdot(4\cdot k-3)}\right)\,\ =\,\ \frac 4\pi$$ or, equivalently, $$\prod_{k=1}^{\infty} \left(1 - \frac 1{(2\cdot k-1)^2\cdot(4\cdot k+1))}\right)\,\ =\,\ \frac \pi 4$$

where I have used the Wallis product. (These equations sound so classical that they must be well-known; I'd appreciate a reference.)

QUESTION   Do you know or can you provide simple proofs of the above equation which do not use the Wallis product?

REMARK 1   Perhaps different proofs of Wallis theorem may lead to different proofs of the above formulas (e.g. via Fourier analysis?).

REMARK 2   In general, products $\ \prod_{k=1}^n (1+a_k)\$ (where $\ a_k\$ are small) can be treated by the Euler's method by studying $\ \sum_{k=1}^n\log(a_k),\$ etc.

• This question got downvoted a couple days ago or so. I am curious what would be the reason? – Włodzimierz Holsztyński Jan 19 '17 at 3:21

## 1 Answer

Let's consider the partial product $$P_n:=\prod_{k=1}^n\frac{(4k+1)(2k-1)^2}{(2k)^2(4k-3)}$$. Then, convert to factorials: \begin{align} P_n&=\prod_{k=1}^n\frac{4k+1}{4k-3}\cdot\prod_{k=1}^n\frac{(2k)^2(2k-1)^2}{(2k)^4} \\ &=\frac{4n+1}{2^{4n}}\binom{2n}n^2=(4n+1)\left(\frac{(2n)!}{2^{2n}n!^2}\right)^2. \end{align} Applying Stirling's approximation $$n!\sim \sqrt{2\pi n}\,\left(\frac{n}e\right)^n$$, we find that \begin{align} P_n \,\,&\sim\,\, (4n+1)\left(\frac{\sqrt{4\pi n}\,\left(\frac{2n}e\right)^{2n}}{2^{2n}(2\pi n)\,\left(\frac{n}e\right)^{2n}} \right)^2 \\ &=\frac{4n+1}{\pi n} \\ &\rightarrow \frac4{\pi}, \end{align} as $$n\rightarrow\infty$$. The proof follows. Note: this does not accelerate the product any more than the Wallis formula does.

If you like to know a bit more about this and related topics, you may look into this paper:

Tewodros Amdeberhan, Olivier R. Espinosa, Victor H. Moll, Armin Straub, Wallis–Ramanujan–Schur–Feynman, American Mathematical Monthly 117 (2010) pp 618–632, doi:10.4169/000298910X496741 , arXiv:1004.2453 (pdf)

• T.A., thank you for your answer. Using Stirling's approximation seems less elementary than using Wallis product--in fact, a popular way to obtain the Stirling's theorem (the version which you use) is by applying the Wallis product, then it's simple. *** Thank you for your link too, I'll have a look. – Włodzimierz Holsztyński Dec 17 '16 at 8:45
• Nice title of that paper, by the way! – theHigherGeometer Aug 20 '19 at 23:57
• @DavidRoberts Thank you! – T. Amdeberhan Aug 21 '19 at 14:23