How to prove the convergence of a series of prime numbers

Question

I have a bit of a problem proving that the series:

$$ \sum_{p\leq x} \frac{p\ln\left(p\right)}{x^2} $$

where the sum is extended over all prime numbers, converges to 0.5.

Any ideas?

Thanks in advance,

Kijn

Answer 1

A standard method is to write the sum as a sum over all positive integers $\leqslant x$, multiplying the term with $\pi(n) - \pi(n-1)$, where $\pi$ is the prime counting function, to annihilate the terms for composite $n$, and then rearrange:

$$\begin{align} \sum_{p\leqslant x} p\log p &= \sum_{n\leqslant x} \bigl(\pi(n)-\pi(n-1)\bigr)n\log n\\ &= \sum_{n \leqslant x} \pi(n)n\log n - \sum_{n\leqslant x-1} \pi(n)(n+1)\log (n+1)\\ &= \pi(x)x\log x - \sum_{n\leqslant x} \pi(n)\left[\log (n+1) + \underbrace{n \log \left(1+\frac{1}{n}\right)}_{1 + O(1/n)}\right] + O(x)\tag{1}\\ &= x^2 - \sum_{n\leqslant x} \underbrace{\pi(n)\log n}_{n + O(n/\log n)} - \underbrace{\sum_{n\leqslant x} \pi(n)}_{O(x^2/\log x)} + O(x^2/\log x)\tag{2}\\ &= x^2 - \sum_{n\leqslant x} n + O(x^2/\log x)\\ &= \frac{x^2}{2} + O\left(\frac{x^2}{\log x}\right). \end{align}$$

Dividing by $x^2$ yields the proposition.

In $(1)$, we use

$$\pi(x)x\log x - \pi\left(\lfloor x\rfloor\right)\lfloor x+1\rfloor \log \lfloor x+1\rfloor \in O(x),$$

and in $(2)$ we use

$$\pi(x) - \frac{x}{\log x} \in O\left(\frac{x}{(\log x)^2}\right)$$

as well as

$$\sum_{2 \leqslant n \leqslant x} \frac{n}{\log n} \in O\left(\frac{x^2}{\log x}\right).$$

At the end,

$$\sum_{n\leqslant x} n = \frac{1}{2} \lfloor x\rfloor\cdot \lfloor x+1\rfloor = \frac{x^2}{2} + O(x).$$