Prime reciprocal lower bound
= Prime reciprocal lower bound
{title2=$\sum_{p\leq x}p^{-1}>\log\log x-1/2$}
For $x\geq2$, the <harmonic number> $H_{\lfloor x\rfloor}$ exceeds $\log x$. The finite <Euler product> over the <primes> at most $x$ contains every summand of this <harmonic number>. Taking <logarithms> and applying the <geometric-series bound for the logarithmic remainder> bounds the total remainder by $\frac12\sum_{n=2}^\infty[n(n-1)]^{-1}=1/2$, proving the displayed bound.