0

For $ n $ be a positive integer and $ f(n) = \dfrac{\pi^2(n)}{n} $, where $ \pi(x) $ is the number of the prime numbers less than $x$ . Then find the value of $\displaystyle S = \lim_{n \to \infty} \dfrac{1}{n} \left(\sum_{k=1}^{n} f(2k)\right).$

By the computation, it seems to be $ S=1 $. But I cannot prove this.

max_zorn
  • 4,875
Makoto Kohno
  • 1
  • I had an important mistake. Sorry. – Makoto Kohno May 27 '18 at 06:09
  • Due to PNT $$\lim_{n \to \infty} \dfrac{1}{n} \left(\sum\limits_{k=1}^{n} f(2k)\right)= \lim_{n \to \infty} \dfrac{1}{n} \left(\sum\limits_{k=1}^{n} \frac{\pi^2(2k)}{2k}\right)\approx \ \lim_{n \to \infty} \dfrac{1}{n} \left(\sum\limits_{k=1}^{n} \frac{2k}{\left(\ln(2k)\right)^2}\right)$$ – rtybase May 27 '18 at 11:01

0 Answers0