= Square-root logarithmic sum over primes
{title2=$\sum_{p\le x}\sqrt{\log p}/p=2\sqrt{\log x}+O(1)$}
Put $A(t)=\sum_{p\le t}(\log p)/p$. The <Mertens first theorem> gives $A(t)=\log t+O(1)$. <Partial summation> yields
$$
\sum_{p\le x}\frac{\sqrt{\log p}}p
=\frac{A(x)}{\sqrt{\log x}}+\frac12\int_2^x\frac{A(t)}{t(\log t)^{3/2}}\,dt.
$$
The main terms sum to $2\sqrt{\log x}$ up to a constant. The error integral is bounded because $\int_2^\infty dt/(t(\log t)^{3/2})$ converges. Thus a <prime sum> with a fractional logarithmic weight follows from an elementary bounded-error estimate, without the <Prime number theorem>.
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