# $\displaystyle\sum@\slimits@@@_{n\leq x}y^{\omega(n)}=O_{y}(x(\mathop{log}% \nolimits x)^{y-1})$ for $y\geq 0$

Within this entry, $\omega$ refers to the number of distinct prime factors function, $\lfloor\,\cdot\,\rfloor$ refers to the floor function, $\log$ refers to the natural logarithm   , $p$ refers to a prime, and $k$ and $n$ refer to positive integers.

###### Theorem 1.

For $y\geq 0$, $\displaystyle\sum_{n\leq x}y^{\omega(n)}=O_{y}(x(\log x)^{y-1})$.

###### Proof.

Since $y^{\omega(p^{k})}=y$ for all $p$ and $k$, the real-valued nonnegative multiplicative function  $y^{\omega(n)}$ the Wirsing condition with $c=y$ and $\lambda=1$. Thus:

 $\displaystyle\sum_{n\leq x}y^{\omega(n)}$ $\displaystyle=O_{y}\left(\frac{x}{\log x}\sum_{n\leq x}\frac{y^{\omega(n)}}{n}\right)$ $\displaystyle=O_{y}\left(\frac{x}{\log x}\prod_{p\leq x}\left(1+\sum_{k=1}^{% \left\lfloor\frac{\log x}{\log p}\right\rfloor}\frac{y^{\omega(p^{k})}}{p^{k}}% \right)\right)$ $\displaystyle=O_{y}\left(\frac{x}{\log x}\left(\exp\left(\sum_{p\leq x}\sum_{k% =1}^{\left\lfloor\frac{\log x}{\log p}\right\rfloor}\frac{y}{p^{k}}\right)% \right)\right)$ $\displaystyle=O_{y}\left(\frac{x}{\log x}\left(\exp\left(y\sum_{p\leq x}\sum_{% k=1}^{\left\lfloor\frac{\log x}{\log p}\right\rfloor}\frac{1}{p^{k}}\right)% \right)\right)$ $\displaystyle=O_{y}\left(\frac{x}{\log x}(\exp(y(\log(\log x)+O(1))))\right)$ $\displaystyle=O_{y}\left(\frac{x}{\log x}(\exp(\log(\log x)^{y}))\right)$ $\displaystyle=O_{y}\left(\frac{x}{\log x}(\log x)^{y}\right)$ $\displaystyle=O_{y}(x(\log x)^{y-1})$

Title $\displaystyle\sum@\slimits@@@_{n\leq x}y^{\omega(n)}=O_{y}(x(\mathop{log}% \nolimits x)^{y-1})$ for $y\geq 0$ displaystylesumnleXYomeganOyxlogXy1ForYge0 2013-03-22 16:11:22 2013-03-22 16:11:22 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Theorem msc 11N37 AsymptoticEstimate DisplaystyleYOmeganOleftFracxlogXy12YRightFor1LeY2 WirsingCondition