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A general infinite product take that form let say, $F(p)=p_1p_2p_3p_4...$

I notices a pattern that accelerate $F(p)$

Accelerate version

$$F(p)=q_1^{1\over 2}q_2^{1\over 4}q_3^{1\over 8}q_4^{1\over 16}...$$

Where

$q_1=p_1$, $q_2=p_1p_2$, $q_3=p_1p_2^2p_3$, $q_4=p_1p_2^3p_3^3p_4$

$q_5=p_1p_2^4p_3^6p_4^4p_5$ And so on .... The power are binomial coefficients

Let apply this idea to Wallis's product

$${\pi\over2}={2\over 1}{2\over3}{4\over3}{4\over5}{6\over5}\cdots$$

$q_1={2\over1}$, $q_2={2\over1}{2\over3}={2^2\over 1\cdot3}$

$q_3={2\over1}{\left(2\over3\right)^2}{4\over3}={2^3\cdot4\over 1\cdot3^3}$

$q_4={2\over1}{\left(2\over3\right)^3}{\left(4\over3\right)^3}{4\over5}={2^4\cdot4^4\over1\cdot3^6\cdot5}$ and son on...

$${\pi\over2}=\left({2\over1}\right)^{1\over2}\left({2^2\over 1\cdot3}\right)^{1\over 4} \left({2^3\cdot4\over 1\cdot3^3}\right)^{1\over8}\left({2^4\cdot4^4\over1\cdot3^6\cdot5}\right)^{1\over16}\cdots$$ Question:

How can I prove that $${\pi\over2}=\left({2\over1}\right)^{1\over2}\left({2^2\over 1\cdot3}\right)^{1\over 4} \left({2^3\cdot4\over 1\cdot3^3}\right)^{1\over8}\left({2^4\cdot4^4\over1\cdot3^6\cdot5}\right)^{1\over16}\cdots$$ is an accelerate of convergence to

$${\pi\over2}={2\over 1}{2\over3}{4\over3}{4\over5}{6\over5}\cdots$$

As in general using the above idea?

gymbvghjkgkjkhgfkl
  • 8,090

1 Answers1

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Take the logarithm of both product representations of $F(p)$ and exchange sums.

$\displaystyle \sum\limits_{k=1}^\infty \ln p_k=\sum\limits_{k=1}^\infty \frac{\ln q_k}{2^k}=\sum\limits_{k=1}^\infty (\ln p_k)\sum\limits_{v=k}^\infty \frac{1}{2^v}\binom{v-1}{k-1}\enspace$ because of $\enspace\displaystyle \ln q_k=\sum\limits_{v=1}^k\binom{k-1}{v-1}\ln p_v$

The reason for the equality is

$$1=[x^k](\frac{1}{1-x})=[x^k](\frac{1}{2}\frac{1}{1-\frac{x+1}{2}})=\sum\limits_{v=k}^\infty \frac{1}{2^v}\binom{v-1}{k-1}$$

with $\enspace k\in\mathbb{N}$.

user90369
  • 11,518