Within this entry, $\tau$ refers to the divisor 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.

The $O_a$ indicates that the constant implied by the definition of $O$ depends on $a$ . (See Landau notation for more details.)

Proof. Let $a \ge 0$ . Since $(\tau)^a={id}^a\circ \tau$ , ${id}$ is completely multiplicative, and $\tau$ is multiplicative, $(\tau)^a$ is multiplicative. (See composition of multiplicative functions for more details.)

For any $y \ge 0$ ,

$\displaystyle \sum_{p \le y} (\tau(p))^a \log p$	$\displaystyle =\sum_{p \le y} 2^a \log p$
	$\displaystyle =2^a\sum_{p \le y} \log p$
	$\displaystyle \le 2^ay\log 4$ by this theorem.

Also,

$\displaystyle \sum_p \sum_{k \ge 2} \frac{(\tau(p^k))^a}{p^k} \log(p^k)$	$\displaystyle = \sum_p \sum_{k \ge 2} \frac{(k+1)^a}{p^k} \cdot k\log p$
	$\displaystyle \le \sum_p \log p \sum_{k \ge 2} \frac{(k+1)^{a+1}}{p^k}$
	$\displaystyle \le \sum_p \frac{\log p}{p^2} \sum_{k \ge 2} \frac{(2k)^{a+1}}{p^{k-2}}$
	$\displaystyle \le 2^{a+1} \sum_p \frac{1}{p^{\frac{3}{2}}} \sum_{k \ge 2} \frac{k^{a+1}}{2^{k-2}}$
	$\displaystyle \le 2^{a+3} \zeta\left( \frac{3}{2} \right) \sum_{k \ge 2} \frac{k^{a+1}}{2^k}$ , where $\zeta$ denotes the Riemann zeta function.

Since

$\displaystyle \sum_{k \ge 2} \frac{k^{a+1}}{2^k}$ converges by the ratio test. Thus, by this theorem, $\displaystyle \sum_{n \le x} (\tau(n))^a=O_a\left(\frac{x}{\log x}\sum_{n \le x} \frac{(\tau(n))^a}{n}\right)$ . Therefore,

$\displaystyle \sum_{n \le x} (\tau(n))^a$	$\displaystyle =O_a\left(\frac{x}{\log x}\sum_{n \le x} \frac{(\tau(n))^a}{n}\right)$
	$\displaystyle =O_a\left(\frac{x}{\log x}\prod_{p \le x}\left( 1+\sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{(\tau(p^k))^a}{p^k} \right) \right)$
	$\displaystyle =O_a\left(\frac{x}{\log x} \left( \exp \left( \sum_{p \le x} \sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{(k+1)^a}{p^k} \right) \right) \right)$
	$\displaystyle =O_a\left(\frac{x}{\log x} \left( \exp \left( \sum_{p \le x} \sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{(2k)^a}{p^k} \right) \right) \right)$
	$\displaystyle =O_a\left(\frac{x}{\log x} \left( \exp \left( 2^a \sum_{p \le x} \sum_{k=1}^{\left\lfloor \frac{\log x}{\log p} \right\rfloor } \frac{k^a}{p^k} \right) \right) \right)$
	$\displaystyle =O_a\left(\frac{x}{\log x}(\exp(2^a(\log \log x+O_a(1)))) \right)$
	$\displaystyle =O_a\left(\frac{x}{\log x}(\exp(\log(\log x)^{2^a})) \right)$
	$\displaystyle =O_a\left(\frac{x}{\log x}(\log x)^{2^a} \right)$
	$\displaystyle =O_a(x(\log x)^{2^a-1})$ .