# $\displaystyle\sum@\slimits@@@_{n\leq x}(\tau(n))^{a}=O_{a}(x(\mathop{log}% \nolimits x)^{2^{a}-1})$ for $a\geq 0$

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 $a\geq 0$, $\displaystyle\sum_{n\leq x}(\tau(n))^{a}=O_{a}(x(\log x)^{2^{a}-1})$.

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

###### Proof.

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

For any $y\geq 0$,

$\displaystyle\sum_{p\leq y}(\tau(p))^{a}\log p$ $\displaystyle=\sum_{p\leq y}2^{a}\log p$ $\displaystyle=2^{a}\sum_{p\leq y}\log p$ $\displaystyle\leq 2^{a}y\log 4$ by this theorem (http://planetmath.org/UpperBoundOnVarthetan).

Also,

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

Since

$\begin{array}[]{ll}\displaystyle\lim_{k\to\infty}\left|\frac{\left(\frac{(k+1)% ^{a+1}}{2^{k+1}}\right)}{\left(\frac{k^{a+1}}{2^{k}}\right)}\right|&% \displaystyle=\lim_{k\to\infty}\left|\frac{(k+1)^{a+1}2^{k}}{k^{a+1}2^{k+1}}% \right|\\ \\ &\displaystyle=\lim_{k\to\infty}\left(\frac{1}{2}\right)\left(\frac{k+1}{k}% \right)^{a+1}\\ \\ &\displaystyle=\frac{1}{2}\left(\lim_{k\to\infty}\frac{k+1}{k}\right)^{a+1}\\ \\ &\displaystyle=\frac{1}{2},\end{array}$

$\displaystyle\sum_{k\geq 2}\frac{k^{a+1}}{2^{k}}$ converges by the ratio test. Thus, by this theorem (http://planetmath.org/AsymptoticEstimatesForRealValuedNonnegativeMultiplicativeFunctions), $\displaystyle\sum_{n\leq x}(\tau(n))^{a}=O_{a}\left(\frac{x}{\log x}\sum_{n% \leq x}\frac{(\tau(n))^{a}}{n}\right)$. Therefore,

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

Title $\displaystyle\sum@\slimits@@@_{n\leq x}(\tau(n))^{a}=O_{a}(x(\mathop{log}% \nolimits x)^{2^{a}-1})$ for $a\geq 0$ displaystylesumnleXtaunaOaxlogX2a1ForAge0 2013-03-22 16:09:53 2013-03-22 16:09:53 Wkbj79 (1863) Wkbj79 (1863) 15 Wkbj79 (1863) Theorem msc 11N37 AsymptoticEstimatesForRealValuedNonnegativeMultiplicativeFunctions DisplaystyleYOmeganOleftFracxlogXy12YRightFor1LeY2