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the lim sup you are asking about is infinite zeraoulia, the article you should read is Alaoglu and Erdos http://www.renyi.hu/~p_erdos/1944-03.pdf The numbers that are most efficient at giving large ( and increasing without bound) $\sigma(n)/n$ are called Colossally Abundant Numbers there. See ...


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If $\sigma(n) < C n$ for a constant $C$ then $\limsup_{n \to \infty} \frac{\sigma(n)}{n \ln \ln n} = 0$. Contradiction.


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No. For example, one has $\frac{\sigma(n!)}{n!} > H_n > \ln(n)$, with $H_n$ being a harmonic number. To prove this, consider that for all $1 \leq k \leq n$ we have $k \mid n!$, so also $\frac{n}{k} \mid n!$. Therefore the sequence $\frac{\sigma(k)}{k}$ contains a subsequence that goes to infinity, so the lim sup is infinity as well, therefore it is ...



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