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Does this inequality have any solutions for composite $n \in \mathbb{N}$?

$$\sqrt{2} < \frac{\sigma_1(n^2)}{n^2} < \frac{4n^2}{(n + 1)^2}$$

Note that $\sigma_1$ is the sum-of-divisors function.

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    $\begingroup$ It seems to hold for most $n$, e.g. for $n=6$, $\sqrt{2}<\sigma_1(36)/36=91/36<144/49$. $\endgroup$ – Zander Jun 9 '13 at 12:17
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Note that,

$\displaystyle f(x)=\frac{4x^2}{(x+1)^2}$ is an increasing function for +ve x(i.e. $x>0$)

$\displaystyle\lim_{x\to \infty }\frac{4x^2}{(x+1)^2}=4$

So $\exists m\in N$ such that $\displaystyle\frac{4x^2}{(x+1)^2}>3$, $\forall x\ge m$

now we will choose an $n=2^a3^b$ with $a,b\ge 1$ with $n\ge m$(Note that one such n exists)

Then we have $\displaystyle\frac{\sigma_{1}(n^2)}{n^2}\le \prod_{i=1}^{k}(\frac{p_i}{p_i-1})=\frac{3}{2}\frac{2}{1}=3<\frac{4n^2}{(n+1)^2}$(The proof of the fact is stated in my observation.)

And we have $n^2=2^{2a}3^{2b}$ as $2a\ge 2$

So $\displaystyle\frac{\sigma_{1}(n^2)}{n^2}=(\sum_{j=0}^{2a}2^{-j})(\sum_{j=0}^{2b}3^{-j})>(1+\frac{1}{2})1=1.5>\sqrt{2}$

Hence there exists one such composite $n$ satisfying the condition of the question.

I have made some obsevations (which might help others in analysing a bit more)which i am stating,

Let $\displaystyle n=\prod_{i=1}^{k}p_i^{\alpha_{i}}$ then we have,

$\displaystyle\sigma_{1}(n^2)=\prod_{i=1}^{k}(\sum_{j=0}^{2\alpha_i}p_{i}^{j})$

Then $\displaystyle\frac{\sigma_{1}(n^2)}{n^2}=\frac{\prod_{i=1}^{k}(\sum_{j=0}^{2\alpha_i}p_{i}^{j})}{\prod_{i=1}^{k}p_i^{2\alpha_{i}}}=\prod_{i=1}^{k}(\sum_{j=0}^{2\alpha_i}p_{i}^{-j})$

And by observing that $p_i^{-j}\le1(\forall 1\le j\le2\alpha_i)$

We have,

$\displaystyle\frac{\sigma_{1}(n^2)}{n^2}=\prod_{i=1}^{k}(\sum_{j=0}^{2\alpha_i}p_{i}^{-j})\le \prod_{i=1}^{k}(2\alpha_i+1)=$no. of divisors of $n^2$

we also have,

$\displaystyle\frac{\sigma_{1}(n^2)}{n^2}=\prod_{i=1}^{k}(\sum_{j=0}^{2\alpha_i}p_{i}^{-j})\le\prod_{i=1}^{k}(\sum_{j=0}^{\infty}p_{i}^{-j})\le \prod_{i=1}^{k}(\frac{1}{1-\frac{1}{p_i}})=\prod_{i=1}^{k}(\frac{p_i}{p_i-1})$

Please feel free to point out mistakes (if there are any).

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  • $\begingroup$ Thank you very much @Abhra Abir Kundu for your detailed answer (for the second time!)... :-) $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 9 '13 at 13:02
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    $\begingroup$ You are welcome..... and thank you for remembering me for my first answer @JoseArnaldoDris $\endgroup$ – Abhra Abir Kundu Jun 9 '13 at 13:13

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