Consider the standard divisor function $\sigma_\alpha(n)$ defined as

$$\sigma_\alpha(n) = \Sigma_{d | n} d^\alpha.$$

Assume that $\alpha > 1$. It was proved at least by Gronwall in [¹] (yeah, this is the same Gronwall who obtained Gronwall’s inequality) that $\sigma_\alpha(n)$ satisfies

$$\sigma_\alpha(n) < \zeta(\alpha) n^\alpha$$

for all $n \in \mathbb{N}$, and this upper bound is achieved. Here $\zeta(\alpha)$ is the Riemann Zeta function.

Recently, thinking with Falko Baustian on one problem, we were partially interested in the asymptotic behaviour of $\sigma_\alpha(n)$ for odd numbers $n$. The natural question is whether we can improve the upper bound for odd $n$’s?

Proposition. Let $\alpha>1$, and let $n \in \mathbb{N}$ be odd. Then

$$\sigma_\alpha(n) < \left(1-\frac{1}{2^\alpha}\right) \zeta(\alpha) n^\alpha,$$

and this upper bound is achieved.

Proof. We will argue sketchely and along the same lines as in [¹], pp. 114-116. First, we decompose an arbitrary $n$ into prime factors as

$$n = p_{\lambda_1}^{\nu_1} \cdot \ldots \cdot p_{\lambda_k}^{\nu_k},$$

where $p_i$ is a prime, $\lambda_i < \lambda_{i+1}$, and $\nu_i > 0$. Note that since $n$ is odd, we have $p_i \neq 2$.

As in formula (5) of [¹], we have

\begin{equation}\label{eq:sigma} \sigma_\alpha(n) = n^\alpha \Pi_{i=1}^k \frac{1-\frac{1}{p_{\lambda_i}^{\alpha(\nu_i+1)}}}{1-\frac{1}{p_{\lambda_i}^\alpha}}. \end{equation}

Since the numerator and denominator in \eqref{eq:sigma} are less than one and since $n$ is odd, we can estimate $\sigma_\alpha(n)$ as

$$\sigma_\alpha(n) < n^\alpha \Pi_{i=1}^k \frac{1}{1-\frac{1}{p_{\lambda_i}^\alpha}} < n^\alpha \Pi_{i=2}^\infty \frac{1}{1-\frac{1}{p_{i}^\alpha}} = n^\alpha \left(1-\frac{1}{2^\alpha}\right) \Pi_{i=1}^\infty \frac{1}{1-\frac{1}{p_{i}^\alpha}} = n^\alpha \left(1-\frac{1}{2^\alpha}\right) \zeta(\alpha),$$

where $p_i$ are consequtive primes $p_1 = 2$, $p_2=3$, etc. Therefore, the desired upper bound is obtained.

---

Let us now prove that the upper bound is achieved. For this end, we consider the admissible sequence

$$n_{k,\nu} = (p_2 \cdot \ldots \cdot p_k)^{\nu-1}.$$

Note that

$$\Pi_{i=2}^k \frac{1-\frac{1}{p_{i}^{\alpha \nu}}}{1-\frac{1}{p_{i}^\alpha}} \to \frac{1-\frac{1}{2^\alpha}}{1-\frac{1}{2^{\nu \alpha}}} \frac{\zeta(\alpha)}{\zeta(\nu \alpha)}$$

uniformly convergent with respect to $\nu$ as $k \to \infty$, see [¹], p. 115. Thus, noting that $1/2^{\nu \alpha} \to 0$ and $\zeta(\nu \alpha) \to 1$ as $\nu \to \infty$, we conclude from \eqref{eq:sigma} that

$$\frac{\sigma_\alpha(n_{k,k+1})}{n_{k,k+1}^\alpha} \to \left(1-\frac{1}{2^\alpha}\right) \zeta(\alpha).$$

---

Bibliography