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I saw somewhere that $L(N) = \sum_{i=2}^N \frac{1}{\log(i)}=\pi(N) (1 + o(1))$ as a consequence of Prime Number Theorem. How to prove this? I think I have to prove that $$\lim_{n \to \infty}\frac{L(N) - \pi(N)}{\pi(N)} = 0$$ using the fact that $\pi(N) \sim \frac{N}{\log(N)}$ but I do not know exactly how ... Or provide a reference ...

C Marius
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  • You have to prove the equivalent result that $$\lim_n \frac{L(n)}{ \pi (n)} = 1$$This is equivalent to $$\lim_n \frac{L(n) \log n}{n}=1$$ – Crostul May 24 '17 at 22:25
  • I know that ... but can you sir provide a more detailed answer ... ? – C Marius May 25 '17 at 07:34

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Your question is not about the PNT but about $\displaystyle\sum_{n=2}^N \frac{1}{\log n} \sim \frac{N}{\ln N}$.

Clearly $$\sum_{n=2}^N \frac{1}{\log n} \sim \int_2^N \frac{dt}{\log t}$$ Let $\displaystyle f(x) = \frac{x}{\ln x}$. Then $\displaystyle f'(x) = \frac{1}{\ln x}-\frac{1}{(\ln x)^2}$.

Thus it is enough to show that $$\int_2^x \frac{dt}{\log t} \sim \int_2^x \frac{dt}{\log t}-\int_2^x \frac{dt}{\log^2 t} \quad (= \frac{x}{\ln x}-\frac{2}{\ln 2})$$ ie. that $$\lim_{x \to \infty}\frac{\displaystyle\int_2^x \frac{dt}{\log^2 t}}{\displaystyle\int_2^x \frac{dt}{\log t}} = 0$$ which is obvious

This leads to the prime number theorem in it useful form :

$$\sum_p p^{-s} \quad \text{ behaves in most case like } \quad \sum_{n=2}^\infty \frac{n^{-s}}{\ln n}$$

reuns
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