Bounds for $\frac{\sigma(q^k)}{2\sigma(q^{k-1})}$ in terms of $q$ and $k$

Question

If $q$ is a prime number and $k$ is a positive integer, does the following double-sided inequality hold?

$$\frac{q}{2} + \frac{q - 1}{2q^k} < \frac{\sigma(q^k)}{2\sigma(q^{k-1})} \leq \frac{q}{2} + \frac{1}{2q^{k-1}}$$

Here, $\sigma(x)$ is the classical sum of divisors of the positive integer $x$.

MY ATTEMPT

Since $\sigma(q^k) = q^k + \sigma(q^{k-1})$, then we have $$\frac{\sigma(q^k)}{2\sigma(q^{k-1})}=\frac{q^k + \sigma(q^{k-1})}{2\sigma(q^{k-1})}=\frac{q}{2}\cdot\frac{q^{k-1}}{\sigma(q^{k-1})}+\frac{1}{2}=\frac{q}{2}\cdot\frac{1}{I(q^{k-1})}+\frac{1}{2},$$ where $I(x)=\sigma(x)/x$ is the abundancy index of $x$.

I also know that $$1 \leq I(q^{k-1}) < \frac{q}{q-1}.$$

Alas, this is where I get stuck.

Answer 1

We have $\dfrac{\sigma(q^k)}{\sigma(q^{k-1})}=\dfrac{q^{k+1}-1}{q^k-1}=q+\color{blue}{\dfrac{q-1}{q^k-1}}$ and $$\frac{q-1}{q^k}<\color{blue}{\dfrac{q-1}{q^k-1}}\leqslant\frac{q-1}{q^k-q^{k-1}}=\frac{1}{q^{k-1}}.$$