How to prove the convergence of a series of prime numbers 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
 A: 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,
$$\sum_{n\leqslant x} n = \frac{1}{2} \lfloor x\rfloor\cdot \lfloor x+1\rfloor = \frac{x^2}{2} + O(x).$$
