How to find an estimate for the sum $\sum_{n\leq x} 2^{\omega(n)}$, where $\omega(n)$ is the number of distinct prime factors of $n$.
Since $2^{\omega(n)}$ is multiplicative, computing its value at prime power, we see that $2^{\omega(n)}=\sum_{d\mid n}\mu^2(d)$. Then
\begin{align}
\sum_{n\leq x}2^{\omega(n)}&=\sum_{n\leq x}\sum_{d\mid n}\mu^2(d)=\sum_{d\leq x}\mu^2(d)\sum_{d\mid n} 1\\
&=\sum_{d\leq x}\mu^2(d)\left\lfloor \frac{x}{d}\right\rfloor=\sum_{d\leq x}\mu^2(d)(\frac{x}{d}+O(1))\\
&=x\sum_{d\leq x}\frac{\mu^2(d)}{d}+O(\sum_{d\leq x}\mu^2(d))
\end{align}
I get stuck here, the series $\sum_{n=1}^\infty\frac{\mu^2(n)}{n}$ is not convergent, I don't know how to estimate the first term.
Best Answer
It is not difficult to see that $$\sum_{d\leq x}\mu^{2}\left(d\right)=x\frac{6}{\pi^{2}}+O\left(\sqrt{x}\right)\tag{1}$$ (for a proof see here) so by Abel's summation we have $$\sum_{d\leq x}\frac{\mu^{2}\left(d\right)}{d}=\frac{\sum_{d\leq x}\mu^{2}\left(d\right)}{x}+\int_{1}^{x}\frac{\sum_{d\leq t}\mu^{2}\left(d\right)}{t^{2}}dt $$ hence using $(1)$ we have $$ \begin{align*} \sum_{d\leq x}\frac{\mu^{2}\left(d\right)}{d}= & \frac{6}{\pi^{2}}+O\left(\frac{1}{\sqrt{x}}\right)+\frac{6}{\pi^{2}}\int_{1}^{x}\frac{1}{t}dt+O\left(\int_{1}^{x}\frac{1}{t^{3/2}}dt\right)\\ = & \frac{6}{\pi^{2}}\log\left(x\right)+O\left(1\right) \end{align*} $$ hence $$\sum_{n\leq x}2^{\omega\left(n\right)}=\frac{6}{\pi^{2}}x\log\left(x\right)+O\left(x\right).$$