Let$\ n$ be an odd abundant number. What is the best upper bound to $\ \frac{\sigma(n)}{n}$? NOTE: Of course the bound of the product of primes below is incorrect. I knew very little of this stuff and it seems was too lazy and gaga (I still am gaga) to check that W|A was acting stupid.
In order to obtain my upper bound I use and extend a method I learnt from Greg Martin, who answered another question of mine.
Lemma: Let $i>1$ be an integer, and let $p$ be the largest prime dividing $i$. Then,
$$
\frac{\sigma(i)}i \le \bigg(1+\frac1p\bigg) \frac{\sigma(i/p)}{i/p}.
$$
Proof: Write $i=p^rm$ where $p\nmid m$. Then,
$$
\frac{\sigma(i)/i}{\sigma(i/p)/(i/p)} = \frac{\sigma(p^rm)}{p\sigma(p^{r-1}m)} = \frac{\sigma(p^r)\sigma(m)}{p\sigma(p^{r-1})\sigma(m)} = \frac{\sigma(p^r)}{p\sigma(p^{r-1})}.
$$
Since $\sigma(p^r) = \frac{p^{r+1}-1}{p-1}$, this gives
$$
\frac{\sigma(i)/i}{\sigma(i/p)/(i/p)} = \frac{p^{r+1}-1}{p(p^r-1)} \le \frac{p+1}p = 1+\frac1p.
$$
Dickson proved there are no odd abundant numbers with less than three prime factors, hence, if$\ P $ is the largest prime dividing$\ n$, we have$\ P\ge7$. So, let$\ n=k\times PA$, where$\ PA$ is any odd primitive abundant number and$\ k$ is an odd integer of the form$\ 3^a 5^b 7^c 11^d...$, where obviously all the exponents are non-negative. We shall use$\ 945$ as$\ PA$ simply because it is the first odd (and primitive) abundant number, but we'll show hereinafter that any$\ PA$ would do fine. Thus, we have $$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(\frac{945k}{7}\right)}{\frac{945k}{7}}.$$ For a start, suppose that in the above factorization of$\ k$, the only positive exponent is$\ a$. Hence,$$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(5\times 3^{a+3}\right)}{5\times 3^{a+3}}=\frac{8\times 6\left(3^{a+4}-1\right)}{7\times 5\times 2\times 3^{a+3}}<\frac{72}{35}.$$ If only$\ a$ and$\ b$ are positive, $$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(3^{a+3}5^{b+1}\right)}{3^{a+3}5^{b+1}}=\frac{8\left(3^{a+4}-1\right)\left(5^{b+2}-1\right)}{7\times 2\times 4\times 3^{a+3}5^{b+1}}<\frac{15}{7}.$$ If only$\ a, b$ and$\ c$ are positive, $$\ \frac{\sigma(n)}{n}\le \frac{8}{7} \frac{\sigma\left(3^{a+3}5^{b+1}7^c\right)}{3^{a+3}5^{b+1}7^c}=\frac{8\left(3^{a+4}-1\right)\left(5^{b+2}-1\right)\left(7^{c+1}-1\right)}{2\times 4\times 6\times 3^{a+3}5^{b+1}7^{c+1}}<\frac{5}{2}.$$ Having reached$\ P$, it is easy to see that from now on as the factorization gets larger by$\ p_n^\alpha$, we just multiply the last ratio by $\ \frac{\left(p_n^{\alpha+1}-1\right)}{p_n^\alpha(p_n-1)}<\frac{p_n}{p_n-1}$, leaving out, aside$\ 2$, only$\ 7$, and this is true for any $\ PA$. 
As a consequence, $$\ \frac{\sigma\left(\frac{n}{7}\right)}{\frac{n}{7}}<\prod_{j=2}^\infty \frac{p_j}{p_j-1}<\frac{11.0454}{2}=5.5227.$$(if it necessary, for the last result see http://www.wolframalpha.com/input/?i=2*+3%2F2+*+5%2F4+*+7%2F6+*+... and click on >More terms in the approximation section)  Therefore, we can conclude $$\ \frac{\sigma(n)}{n}<\frac{8}{7} 5.5227=6.3116\overline {571428}.$$

Ummm...$\frac{\sigma(n)}{n}$ is unbounded above for odd  $n$.
Lemma: For all $n\in\mathbb{N}$, we have $\frac{\sigma(n)}{n}=\sum_{d\mid n} \frac{1}{d}$.
Proof: Let $d_1<d_2<\dots<d_t$ be the divisors of $n$. Then $\frac{\sigma(n)}{n}=\frac{d_1}{n}+\cdots+\frac{d_t}{n}=\frac{1}{d_t}+\cdots+\frac{1}{d_1}$.  QED.
Now if we choose an odd $n$ that is divisible by $3, 5, 7, \dots, 2k+1$, then $\frac{\sigma(n)}{n}\ge \frac{1}{3}+\frac{1}{5}+\frac17+\cdots+\frac{1}{2k+1}$.  This is a partial sum of the odd harmonic series, which is divergent.  So by choosing $k$ large enough, we can make $\frac{\sigma(n)}{n}$ as large as we please.
