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While self studying analytic number theory from Introduction to sieve methods and it's applications by M Ram Murthy and Alina Carmen,I have a doubt in theorem 1.4.1 proof by Chebysheff.

My doubt is -> Author writes $\frac {(2n) ! } { ( n!) ^2} \leq 2^{2n} $ and then he writes this step which I am not able to derive - upon taking logarithms $\theta(2n) - \theta(n) \leq 2n log 2 $

Can someone please help in how to derive this statement!!

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  • $\begingroup$ What is $\theta$? $\endgroup$ – cangrejo Jan 15 at 12:13
  • $\begingroup$ $\theta(x) = \sum_{p \leq x} ln( p ) $ , Its 2nd chebycheff function $\endgroup$ – Ben Jan 15 at 12:15
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    $\begingroup$ I think there is a mistake. It should be $\theta(2n)-2\theta(n)$. $\endgroup$ – cangrejo Jan 15 at 12:29
  • $\begingroup$ @broncoAbierto I checked the book again . the statement mentioned by me in question is correct $\endgroup$ – Ben Jan 15 at 12:41
  • $\begingroup$ What do you mean correct? If the book says that, I believe it is a typo. I checked numerically and my derivation below checks out. $\endgroup$ – cangrejo Jan 15 at 12:45
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We have $\frac{(2n)!}{(n!)^2}=\binom{2n}{n}\in \mathbb{N}$. Therefore $p|\binom{2n}{n}$ for all primes $n<p<2n$ and $\prod_{n<p<2n}p|\binom{2n}{n}<2^{2n}$. Now use $\ln$ to get the desired inequality.

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  • $\begingroup$ can you please look at this question if you have some time to spare. I am badly struck on it. Perhaps its not right way to ask but I really need help.math.stackexchange.com/questions/3523183/… $\endgroup$ – Ben Feb 3 at 6:04

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