# Proof of Ramanujan's identity

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.

• What is $\psi$? – Wojowu Aug 16 '15 at 20:27
• It's the Chebyshev function, psi. Sorry, I forgot to mention that. – maciek45 Aug 16 '15 at 20:33
• the definition of the function is $\psi(x)=\sum_{p\le x} k \log p$. – maciek45 Aug 16 '15 at 20:34
• What is k in your def'n of Chebyshev function? – DanielWainfleet Aug 16 '15 at 21:02
• 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 – maciek45 Aug 16 '15 at 21:15

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

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

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

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