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

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

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

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

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

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

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

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

Note that

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

# Bibliography

1. Gronwall, T. H. (1913). Some asymptotic expressions in the theory of numbers. Transactions of the American Mathematical Society, 14(1), 113-122.  2 3 4