Within this entry, $\Omega$ refers to the number of (nondistinct) prime factors function, $\mu$ refers to the Möbius function, $\log$ refers to the natural logarithm, $p$ refers to a prime, and $d$ $k$ $m$ and $n$ refer to positive integers.

Proof. Let $g$ be a function such that $y^{\Omega}=1*g$ Then $g$ is multiplicative and $g=\mu*y^{\Omega}$ Thus:

$\displaystyle \sum_{n \le x} y^{\Omega(n)}$	$\displaystyle =\sum_{d \le x} \sum_{m \le \frac{x}{d}} g(d)$ by the convolution method
	$\displaystyle =O\left( \sum_{d \le x} g(d) \cdot \frac{x}{d} \right)$
	$\displaystyle =O\left( x \prod_{p \le x} \left( 1+\sum_{k} \frac{g(p^k)}{p^k} \right) \right)$
	$\displaystyle =O\left( x \prod_{p \le x} \left( 1+\sum_{k} \frac{\mu(1)y^k+\mu(p)y^{k-1}}{p^k} \right) \right)$
	$\displaystyle =O\left( x \prod_{p \le x} \left( 1+\frac{y-1}{p} \sum_{k} \left( \frac{y}{p} \right)^{k-1} \right) \right)$
	$\displaystyle =O\left( x \prod_{p \le x} \left( 1+\frac{y-1}{p} \cdot \frac{1}{1-\frac{y}{p}} \right) \right)$
	$\displaystyle =O\left( x \prod_{p \le x} \left( 1+\frac{y-1}{p-y} \right) \right)$
	$\displaystyle =O\left( x \left(1+\frac{y-1}{2-y} \right) \prod_{3 \le p \le x} \left( 1+\frac{y-1}{p-y} \right) \right)$
	$\displaystyle =O\left( x \left(\frac{2-y+y-1}{2-y} \right) \prod_{3 \le p \le x} \exp \left(\frac{y-1}{p-y} \right) \right)$
	$\displaystyle =O\left( \frac{x}{2-y} \left( \exp \left(\sum_{3 \le p \le x} \frac{y-1}{p-y} \right) \right) \right)$
	$\displaystyle =O\left( \frac{x}{2-y} \left( \exp \left( \sum_{3 \le p \le x} \frac{1}{p-y} \right) \right)^{y-1} \right)$
	$\displaystyle =O\left( \frac{x}{2-y} \left( \exp \left( \log\log x+O(1) \right) \right)^{y-1} \right)$
	$\displaystyle =O\left( \frac{x}{2-y} \left(ce^{\log\log x} \right)^{y-1} \right)$ for some $c>0$
	$\displaystyle =O\left( \frac{x}{2-y} \left(\max\{1,c\}\log x\right)^{y-1} \right)$
	$\displaystyle =O\left( \frac{x}{2-y} \left(\max\{1,c\}\right)^{2-1} \left( \log x \right)^{y-1} \right)$
	$\displaystyle =O\left( \frac{x \left( \log x \right)^{y-1}}{2-y} \right)$

Note that a similar result for $y=2$ (and therefore for $y \ge 2$ , such as $\displaystyle \sum_{n \le x} 2^{\Omega(n)}=O(x\log x)$ is unobtainable, as evidenced by this theorem. On the other hand, the asymptotic estimates $\displaystyle \sum_{n \le x} 2^{\omega(n)}=O(x\log x)$ and $\displaystyle \sum_{n \le x} \tau(n)=O(x\log x)$ are true.