$x{log}^{2}x=O\left({\displaystyle \sum \mathrm{@}\mathrm{@}\mathrm{@}{\mathrm{@}}_{n\le x}{2}^{\mathrm{\Omega }(n)}}\right)$

Within this entry, $\mathrm{\Omega }$ refers to the number of (nondistinct) prime factors function (http://planetmath.org/NumberOfNondistinctPrimeFactorsFunction), $\tau$ refers to the divisor function, $\mu$ refers to the Möbius function, $\lfloor \cdot \rfloor$ refers to the floor function, $\log$ refers to the natural logarithm, $p$ refers to a prime, and $d$, $k$, $\mathrm{\ell }$, $m$, and $n$ refer to positive integers.

Theorem.

$x{\log }^{2}x=O\left({\displaystyle \sum _{n\le x}}{2}^{\mathrm{\Omega }(n)}\right)$

Proof.

Since, for $x$ sufficiently large, $\log x=O\left(\log \left({\displaystyle \frac{x}{2}}\right)\right)$ and $\log x=O\left(\lfloor {\displaystyle \frac{\log x}{\log 2}}\rfloor \right)$, it follows that $x{\log }^{2}x=O\left({\displaystyle \sum _{n\le x}}{2}^{\mathrm{\Omega }(n)}\right)$.∎