TITLE: Density of outputs of divisor function
QUESTION [6 upvotes]: We define the function:
$$\sigma_k{(n)}=\sum_{d \mid n} d^k$$
as the sum of divisors function of the $k$th power, where $k \in \mathbb{N_0}$, for $n \in \mathbb{N}$. Now, we define:
$$S_k=\{a \space | \space \exists \space m \in \mathbb{N} \space ; \sigma_k{(m)}=a\}$$
It is clear that $S_0$ is the same as $\mathbb{N}$ since for any $n \in \mathbb{N}$, we have $\sigma_0{(2^{n-1})}=n$. However, it is clear than this is not true for $S_k$ where $k>0$.

What is the density of $S_k$ in $\mathbb{N}$ for $k \in \mathbb{N}$?

Edit: As pointed out by @ThomasAndrews in the comments below, $k>1$ has zero density by bounding. Now, the problem is solved if anybody is able to find the density of the sum of divisors function $\sigma(x)$ ($1$st power).

REPLY [2 votes]: This result can be found in the paper R. Sita Rama Chandra Rao and G. Sri Rama Chandra Murty: On a Theorem of Niven, Canad. Math. Bull. 22 (1979), no. 1, 113–115. DOI: 10.4153/CMB-1979-018-5, Zbl: 0401.10069, MR532280
The paper relies on some results from the paper Ivan Niven: The asymptotic density of sequences. Bull. Amer. Math. Soc., 57(6):420-434, 1951; Zbl: 0044.03603, MR0044561.

The proof $k=1$ is short enough to be reproduced here.
We want to show that the set
$$C=\{\sigma(n); n\in\mathbb N\}$$
has zero density.
We will denote
\begin{align*}
d(A)&=\lim_{n\to\infty} \frac{A(n)}n\\
\overline d(A)&=\limsup_{n\to\infty} \frac{A(n)}n\\
\end{align*}
for $A\subseteq \mathbb N$, where $A(n)=|\{k\in A; k\le n\}|$ denotes the number of the element os $A$ which are less than or equal to $n$. I.e, $d(A)$ denotes the asymptotic density and $\overline d(A)$ denotes the upper asymptotic density. (You can notice that $d(A)$ is defined only for some sets, while $\overline d(A)$ exists for each set. However, if $\overline d(A)=0$, then we get also $d(A)=0$. It is also useful to notice that $\overline d(A\cup B)\le \overline d(A)+\overline d(B)$.)
Let us fix an integer $k$. We divide the set $C$ into
\begin{align*}
C_1&=\{s\in C; 2^k\mid C\}\\
C_2&=\{s\in C; 2^k\nmid C\}
\end{align*}
It is clear that $\overline d(C_1)\le\frac1{2^k}$.
If $s\in C_2$, then $s=\sigma(t)$ for some $t$ which has at most $k$ prime factors with exponent one.1 Moreover, we have $t\le s$. 
Let $D$ be the of all positive integers which have at most $k$ primes factors with exponent one. We see that $C_2(n)\le D(n)$, since if $s\in C_2$ and $s\le n$, then for the corresponding $t\in D$ we also have $t\le n$.
Since $d(D)=0$ (Niven, Corollary 2), we get
$$d(C_2)=\lim_{n\to\infty} \frac{C_2(n)}n \le \lim_{n\to\infty} \frac{D(n)}n=0.$$
Consequently, we have
$$\overline d(C)\le \overline d(C_1)+\overline d(C_2) = \frac1{2^k}.$$
Since the inequality $\overline d(C)\le\frac1{2^k}$ is true for every $k$, we get $\overline d(C)=0$ and, consequently, $d(C)=0$.
1Recall that if $n=p_1^{a_1}p_2^{a_2}\dots p_t^{a_t}$, then $\sigma(n)=\prod\limits_{i=1}^t\sigma(p_i^{a_i})$. If $a_i=1$ and $p_i$ is odd, then $\sigma(p_i^{a_i})=p_i+1$ is even. So if $n$ contains at least $k+1$ prime factors in first power, then at least $k$ of them is odd and we get $2^k\mid \sigma(n)$.