Global extrema for $\frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)}$?

Question

Does the function $$\frac{(q^k-1)(q^{k+1}-2q^k+1)}{{q^k}(q-1)(q^{k+1}-1)}$$ have either a global minimum or a global maximum for $q \geq 5$ and $k \geq 1$?

WolframAlpha is unable to find any.

WolframAlpha computation for the global minimum

WolframAlpha computation for the global maximum

Answer 1

can we do better than this?

No, we cannot.

Let $f(k)$ be your function.

Then, we have $$f'(k)=\frac{ (q-4)q^{2 k + 1} +2 q^{k + 1} + 2 q^{2 k} - 1}{q^k(q - 1) (q^{k + 1} - 1)^2}\ln q$$ which is positive for $k\ge 1$ and $q\ge 5$.

So, we see that $f(k)$ is increasing for $k\ge 1$.

Since $$f(1)=\frac{q-1}{{q}(q+1)},\qquad \lim_{k\to\infty}f(k)=\lim_{k\to\infty}\frac{(1-\frac{1}{q^k})(1-\frac{2}{q}+\frac{1}{q^{k+1}})}{(q-1)(1-\frac{1}{q^{k+1}})}=\frac{q-2}{q(q-1)}$$ we have $$\frac{q-1}{{q}(q+1)}\le f(k)\lt \frac{q-2}{q(q-1)}$$

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Added :

Let $f(q)$ be your function.

We get, with some help of WA, $$f'(q)=\frac{- (q^{k + 1} - k q + k - q) (q^{2k+1}(q-4)+2q^{k+1}+2q^{2k}-1)}{q^{k+1}(q-1)^2(q^{k+1}-1)^2}$$

Here, let $$g(q):=q^{k + 1} - k q + k - q$$Then,$$g'(q)=(q^k-1)(k+1)\gt 0$$So, $g(q)$ is increasing and $$g(q)\ge g(5)=5^{k + 1} - 4k - 5\gt 0$$

It follows that $f(q)$ is strictly decreasing for $q\ge 5$.

Since $\displaystyle\lim_{q\to\infty}f(q)=0$, we have $$0\lt f(q)\le f(5)=\frac{(5^k-1)(5^{k+1}-2\cdot 5^k+1)}{4\cdot 5^k(5^{k+1}-1)}\lt \frac{3}{20}$$