I was reading the following paper. But I can't understand why the last line concerning $\frac{2}{\pi}$ is true. The proof is a work of Sylvester. I would be happy if someone helps me in understanding why the last inequality follows.

Thanks in advance.

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As Fedor Petrov says, this looks incorrect. The presence of

$$2/\pi = \prod_{n}\left(1-\frac{1}{(2n)^2}\right)$$

(known as the Wallis product) makes me think Sylvester is mistakenly comparing it to this somehow, with the missing implicit step being 'every prime is at least some even number', so $M\geq W$, since if $2n\leq p$ then we can replace the $(1-1/(2n)^2)$ factor with $(1-1/p^2)$ and only increase the product.

The error is, of course, that 'the greatest even number $\leq p$' is not unique for each $p$ - one gets the same factor appearing for both $2$ and $3$! If one corrects this by omitting $3$ then you get

$$ M > (1-1/9)\cdot\frac{2}{\pi} = \frac{16}{9\pi}.$$

I don't know if this correction is enough to salvage the remainder of Sylvester's argument. (Note that this is now consistent with the fact that $\prod_p (1-1/p^2)=6/\pi^2$).

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  • $\begingroup$ Thanks for the explanation. $\endgroup$ – math is fun Jun 5 at 9:56
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    $\begingroup$ Yeah. This correction is now enough to salvage Sylvester's argument. Actually we needed only $ M> x$ for some $x$ where $0<x< \infty$. Thanks a lot. $\endgroup$ – math is fun Jun 5 at 9:59

I also do not understand. The infinite product of $(1-1/p^2)$ equals $6/\pi^2<2/\pi$.

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  • $\begingroup$ Thanks you for the answer. $\endgroup$ – math is fun Jun 5 at 9:56
  • $\begingroup$ I think this result was due to Euler, who invented it by the time this paper was published ! $\endgroup$ – adityaguharoy Jun 24 at 16:56

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