Is $\frac{\sigma(q^k)}{\sigma(n^2)}$ bounded from below by a function of $n$, if $q \equiv 1 \pmod 4$ is prime and $n^2$ is deficient?

If $q$ is a prime with $q \equiv k \equiv 1 \pmod 4$ and $n^2$ is deficient, is $\frac{\sigma(q^k)}{\sigma(n^2)}$ bounded from below by a function of $n$?

Of course, an easy lower bound is $$\frac{\sigma(q^k)}{\sigma(n^2)} \geq \frac{q^k + 1}{\sigma(n^2)} > \frac{q^k + 1}{2n^2} \geq \frac{5^1 + 1}{2n^2} = \frac{3}{n^2}.$$

I am interested in lower bounds of the form $$\frac{\sigma(q^k)}{\sigma(n^2)} \geq \frac{a}{bn},$$ where $a, b \in \mathbb{R}$.

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$\sigma_1(q^k)$ is constant wrt $n$ I presume, so this is essentially finding a $c$ such that $\sigma_1(n^2)\le cn$. – anon Jul 1 '12 at 23:53
Yes anon, in a way that is correct... – Arnie Bebita Dris Jul 1 '12 at 23:58
Well, of course $\sigma_1(n^2)$ will be bounded by a function of $n$ (itself, namely. However, $\sigma_1(n^2)\ge n\sigma_1(n)$, and $\sigma_1$ does not grow as a constant function, so $\sigma_1(n^2)$ is beyond linear as a function of $n$ and no $c$ will work. – anon Jul 2 '12 at 0:13
Thanks for that comment, anon! Can you write it as an answer, so that I may be able to accept it accordingly? :) – Arnie Bebita Dris Jul 2 '12 at 11:00

Functions of the form $\displaystyle \sum_{d|n}f(d)$ are multiplicative, so in particular $\sigma_1$ is. This gives rise to:
$$\sigma_1(n)=\prod_{p^r\|n}\frac{p^{r+1}-1}{p-1}>\prod_{p^r\|n}\frac{p^{r+1}-p^r}{p-1}=\prod p^r=n, \tag{n>1}$$
where $p^r\|n$ means $p^r$ is the highest power of the prime $p$ dividing $n$. Hence $\sigma_1(n^2)>n^2$ grows not only above any linear function, but above $n^2$. However Grönwall's theorem says asymptotically that $\sigma_1(n^2)$ shouldn't grow any faster than $e^\gamma n^2\log (\log(2n))$ (on the Riemann Hypothesis, this should hold for all sufficiently large $n$), where $\gamma$ is the Euler-Mascheroni constant. I'm not sure if stipulating $n$ is deficient easily lends to any reasonably better bounds.