How do i show this :$\lim_{k\to\infty} \frac{\sigma_{2k+1}(n)}{\sigma_{2k-1}(n)}=n²$ if it is true?

Question

I run some computation in wolfram alpha I find for many fixed values of $n$ and for an arbitrary integer $k$ the ratio :

$\frac{\sigma_{2k+1}(n)}{\sigma_{2k-1}(n)}$ close to $n²$ .

My question here : How do i show that :

$$\lim_{k\to\infty}\frac{\sigma_{2k+1}(n)}{\sigma_{2k-1}(n)}=n²$$

Thank you for any help !!!