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I'm having trouble understanding Ramanujan's formula from his proof of Bertrand's postulate, namely: $$ \ln \lfloor x\rfloor!=\sum_{k=1}^{\infty}\psi\left(\frac{x}{k}\right) $$ where $ \ln x = \log_ex$. Could someone explain me step by step, how to prove the formula? Thank you in advance.

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  • $\begingroup$ What is $\psi$? $\endgroup$ – Wojowu Aug 16 '15 at 20:27
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    $\begingroup$ It's the Chebyshev function, psi. Sorry, I forgot to mention that. $\endgroup$ – maciek45 Aug 16 '15 at 20:33
  • $\begingroup$ the definition of the function is $ \psi(x)=\sum_{p\le x} k \log p $. $\endgroup$ – maciek45 Aug 16 '15 at 20:34
  • $\begingroup$ What is k in your def'n of Chebyshev function? $\endgroup$ – DanielWainfleet Aug 16 '15 at 21:02
  • $\begingroup$ I'm quoting Wikipedia: the second Chebyshev function $ \psi(x) $ is defined similarly, with the sum extending over all prime powers not exceeding x: $ \psi(x) = \sum_{p^k\le x}\log p $. Sorry, I misstated the definition. Thanks for patience ;D $\endgroup$ – maciek45 Aug 16 '15 at 21:15
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For this proof we will use the definition

$$\psi(x) = \sum_{n \le x} \Lambda(n)$$

and the identity

$$\log n = \sum_{d|n} \Lambda(n)$$

which is proven here.

Now

$$\log [x]! = \sum_{n \le x} \log n = \sum_{n \le x} \sum_{d|n} \Lambda(d)$$

$$ = \sum_{ed=n} \Lambda(d) = \sum_{e \le x} \sum_{d \le x/e} \Lambda(d)$$

$$ = \sum_{e \le x} \psi(x/e).$$

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