${\displaystyle \sum \mathrm{@}\mathrm{@}\mathrm{@}{\mathrm{@}}_{n\le x}{(\tau (n))}^a}=O_a(x{(logx)}^{2^a-1})$ for $a\ge 0$

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.

Theorem.

For $a\mathrm{\ge }\mathrm{0}$, ${\displaystyle \mathrm{\sum }_{n\mathrm{\le }x}}{\mathrm{(}\tau \mathit{}\mathrm{(}n\mathrm{)}\mathrm{)}}^a\mathrm{=}{O}_a\mathit{}\mathrm{(}x\mathit{}{\mathrm{(}\log \mathit{}x\mathrm{)}}^{{\mathrm{2}}^a\mathrm{-}\mathrm{1}}\mathrm{)}$.

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

For any $y\ge 0$,

Also,

Since

${\displaystyle \sum _{k\ge 2}}{\displaystyle \frac{k^{a+1}}{2^k}}$ converges by the ratio test. Thus, by this theorem (http://planetmath.org/AsymptoticEstimatesForRealValuedNonnegativeMultiplicativeFunctions), ${\displaystyle \sum _{n\le x}}{(\tau (n))}^a=O_a\left({\displaystyle \frac{x}{\log x}}{\displaystyle \sum _{n\le x}}{\displaystyle \frac{{(\tau (n))}^a}{n}}\right)$. Therefore,

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