In this paper, Erdös presents the following claim:
For every $k$ and $\delta>0$ and for all $n$ except a sequence of density $0$, $$(1-\delta)n\Big{(}\frac{s(n)}{n}\Big{)}^i<s^i(n)<(1+\delta) n\Big{(}\frac{s(n)}{n}\Big{)}^i \tag{1}$$ for $1\leq i\leq k$
Where $s(n)=s^1(n)=(\sum_{d|n}d)-n$, and $s^i(n)=s(s^{i-1}(n))$ is the $i^{th}$ term of the aliquot sequence starting at $n$.
I do not understand the role which $k$ plays in the statement above. Why can $(1)$ not be rephrased as $$(1-\delta)n\Big{(}\frac{s(n)}{n}\Big{)}^i<s^i(n)<(1+\delta) n\Big{(}\frac{s(n)}{n}\Big{)}^i$$ for $\textbf{all}$ $i$? How does the upper bound on $i$ by $k$ impact the statement? Additionally, given that it is known that abundant numbers (defined as $s(n)>n$) have nonzero density, would the claim above not predict that the aliquot sequence of an abundant number grows without bound?
I'm clearly misinterpreting Erdös' claim. So what is it really saying?
What you're missing is that while the union of finitely many sets of density $0$ in $\mathbb N$ has density $0$, the union of countably many might not. A singleton $\{i\}$ has density $0$, but $\mathbb N = \bigcup_{i\in \mathbb N} \{i\}$.
Thus let $B(i, \delta)$ be the set of $n$ for which (1) is false. Erdös's claim is equivalent to saying that for each $i$ and $\delta$, $B(i,\delta)$ is a set of density $0$.
Then if you take any $k$, (1) holds for $0 \le i \le k$ if $n$ is not in $\bigcup_{i=0}^k B(i,\delta)$, which is a set of density $0$.
But if you're asking for (1) to hold for all $i \ge 0$, you need $n$ to not be in $\bigcup_{i=0}^\infty B(i,\delta)$, and that might not be a set of density $0$.