Asymptotics for series

by Wolker   Last Updated December 07, 2017 10:20 AM

What can be said about an equivalent of $$\sum_{n \leqslant x} \frac{n}{\log n}$$

I would like to compare it to $x^2$ which is approximately $\sum_{n \leqslant x} n$. Is it negligible in front of it? I tried partial summation and dyadic cutting, but nothing seems to work...

Answers 1

According to prime number theorem $$\lim_{xto\infty}\dfrac{\pi(x)\log x}{x}=1$$ so $$\dfrac{x}{\log x}\sim\pi(x)$$ where $\pi(x)$ is number of prime numbers less than $x$.

December 07, 2017 10:18 AM

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