# Limit involving primes: $\lim_{n\to\infty}\frac{1}{n}\sum_{k = 1}^{n}\left(\frac{1}{2}-\frac{1}{4}\frac{\log \log p_k }{\log p_k} \right)$

I have the following sum that I am trying to put into a cleaner formula (one that I can hopefully find the value of). This looks like it may be similar to a Riemann Sum, so it could turn into an integral. Any ideas on how to evaluate it? $$\lim_{n\to\infty}\frac{1}{n}\sum_{k = 1}^{n}\left(\frac{1}{2}-\frac{1}{4}\frac{\log \log p_k }{\log p_k} \right)$$

$p_k$ is the $k^{th}$ prime.

• Your limit is $1/2$ because $\frac{\log \log p_k}{\log p_k} \to 0$. Also the prime number theorem says $p_k \sim k \log k$ and $\sum_{k=1}^n f(p_k) \sim \sum_{k=2}^{n \log n} \frac{f(k)}{\log k}$ with $f(x) = \frac{\log \log x}{\log x}$ – reuns Aug 16 '17 at 1:36
• @reuns p_k is the k-th prime, not the pi function. Also, how were you able to get rid of the fraction that approaches zero if it's inside the sum? Thanks! – Eric Miller Aug 28 '17 at 22:40
• (never mind that first sentence in the above comment) – Eric Miller Aug 28 '17 at 22:51
• If $\lim_{k \to \infty} a_k = 0$ then $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k = ?$ – reuns Aug 29 '17 at 12:02
• @reuns can you link me to a source for that (I know it's right but I need a source). Thanks again! – Eric Miller Aug 29 '17 at 18:07