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. 
 A: 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)

