# Let $k\in\mathbb{N},$ Show that there is $i\in\mathbb{N}$ s.t $\ \left(1-\frac{1}{k}\right)^{i}-\left(1-\frac{2}{k}\right)^{i}\geq \frac{1}{4}$

let $$k\in\mathbb{N},$$ Show that there is a choice of $$i\in\mathbb{N}$$ such that

$$\left(1-\frac{1}{k}\right)^{i}-\left(1-\frac{2}{k}\right)^{i}\geq \frac{1}{4}.$$

Things I tried:

1. $$i=k,$$ doesn't hold for any $$k,$$
2. I tried to use online calculators, but it seems the answer is some function of $$k$$. I can't tell which exactly.
3. I tried to use Bernoulli's inequality and related inequality for the left and right expression:

$$(1 - x)^n \geq 1 - nx$$

$$-(1 - x)^n \geq -e^{-xn}$$

Each one makes the expression smaller than $$\frac{1}{4}$$ for any $$i$$ or $$k.$$

1. I tried to use the Newton's Binomial Theorem and didn't succeed to get anything for that.
2. I thought about maybe using Taylor expansion, but I don't have any idea how to do it.
• Really, that seems wrong. Given a $i$, there is always a $k$ such that $1-(1-\frac 2 k)^i < \frac 1 4$. Are you sure $i$ does not depend upon $K$? Commented Aug 5, 2023 at 14:46
• I also think it is wrong. $i=1$ doesn't work, because $\left(1-\frac{1}{k}\right)^{1}-\left(1-\frac{2}{k}\right)^{1} = \frac{1}{k},$ so this fails for $k = 5.$ Next, for $k=1$ to work, $i$ must be odd. But if $i$ is odd and $i>1,$ then $k=2$ never works. Commented Aug 5, 2023 at 14:50
• fixed, instructor wording was not 100% clear, but all of my tryouts are still hold. Commented Aug 5, 2023 at 15:42
• I tried using desmos, and I got an interesting result. Considering $f(x)=\left(1-\dfrac{1}{x} \right) ^i - \left(1-\dfrac{2}{x} \right) ^i -\dfrac{1}{4}$, $f(x) \ge 0$ only when $i \approx \ln(2) x - \dfrac{1}{2}$ and $f(x)$ has an asymptote $y=0$ when $i = \ln(2) x - \dfrac{1}{2}$ Commented Aug 5, 2023 at 19:59

## 1 Answer

Edit: Okay, I added the details below. I missed this in my original skim of the argument, but the (k=3) is most easily handled as an exception as well. To make (my) life easier, I've slightly altered some of the notation. (For instance, $$\log=\ln$$)

Original post: Not a full answer, but here’s a sketch (the $$k=1$$ case is invalid, the $$k=2$$ case is trivial, and the $$k=3$$ case is easy):

1. Show that for every $$k\in\mathbb{N}_{\geq 4}$$, the function $$f_{k}\left(x\right)\ :=\ \left(1-\tfrac{1}{k}\right)^{x}-\left(1-\tfrac{2}{k}\right)^{x}$$ defined for $$x\in\left(-\infty,\infty\right)$$ attains a unique local maximum, before which it is strictly increasing, and after which it is strictly decreasing. It follows that on any closed subinterval of $$\left[1,\infty\right)$$, $$f_{k}$$ is minimized (only) at (one or both of) the endpoints.

2. Show that the function $$\alpha_{0}\left(k\right)\ :=\ \underbrace{\left(1-\tfrac{1}{k}\right)^{\log\left(2\right)k}}_{:=\ \alpha_{0;1}\left(k\right)}\ -\ \underbrace{\left(1-\tfrac{2}{k}\right)^{\log\left(2\right)k}}_{:=\ \alpha_{0;2}\left(k\right)}$$ defined for $$k\in\left[4,\infty\right)$$ is strictly decreasing and attains a limit of $$\tfrac{1}{4}$$ as $$k\to\infty$$.

3. Show that the function $$\alpha_{1}\left(k\right)\ :=\ \underbrace{\left(1-\tfrac{1}{k}\right)^{\log\left(2\right)k-1}}_{:=\ \alpha_{1;1}\left(k\right)}\ -\ \underbrace{\left(1-\tfrac{2}{k}\right)^{\log\left(2\right)k-1}}_{:=\ \alpha_{1;2}\left(k\right)}$$ defined for $$k\in\left[4,\infty\right)$$ is strictly decreasing and attains a limit of $$\tfrac{1}{4}$$ as $$k\to\infty$$.

4. Conclude that for any $$k\in\mathbb{N}_{\geq 4}$$, \begin{align*} \left(1-\tfrac{1}{k}\right)^{\left\lfloor \log\left(2\right)k\right\rfloor} - \left(1-\tfrac{2}{k}\right)^{\left\lfloor \log\left(2\right)k\right\rfloor}\ &>\ \min\left(\left(1-\tfrac{1}{k}\right)^{\log\left(2\right)k} - \left(1-\tfrac{2}{k}\right)^{\log\left(2\right)k-1},\ \left(1-\tfrac{1}{k}\right)^{\log\left(2\right)k-1} - \left(1-\tfrac{2}{k}\right)^{\log\left(2\right)k-1}\right)\\ &>\ \tfrac{1}{4} \end{align*} so that $$\left\lfloor \log\left(2\right)k\right\rfloor$$ is the desired natural.

Proof of Claim 1: Indeed, for $$k>2$$ we have that \begin{align*} \frac{\text{d}f_{k}}{\text{d}x}\left(x\right)\ >\ 0\ &\iff\ \log\left(1-\tfrac{1}{k}\right)\exp\left(\log\left(1-\tfrac{1}{k}\right)x\right)-\log\left(1-\tfrac{2}{k}\right)\exp\left(\log\left(1-\tfrac{2}{k}\right)x\right)\ >\ 0\\ &\iff\ \log\left(-\log\left(1-\tfrac{2}{k}\right)\right)\ +\ \log\left(1-\tfrac{2}{k}\right)x\ >\ \log\left(-\log\left(1-\tfrac{1}{k}\right)\right)\ +\ \log\left(1-\tfrac{1}{k}\right)x\\ &\iff\ \frac{\log\left(-\log\left(1-\tfrac{2}{k}\right)\right)-\log\left(-\log\left(1-\tfrac{1}{k}\right)\right)}{\log\left(1-\tfrac{1}{k}\right)-\log\left(1-\tfrac{2}{k}\right)}\ >\ x \end{align*} and by a symmetric argument that $$\frac{\text{d}f_{k}}{\text{d}x}\left(x\right)\ <\ 0\ \iff\ \frac{\log\left(-\log\left(1-\tfrac{2}{k}\right)\right)-\log\left(-\log\left(1-\tfrac{1}{k}\right)\right)}{\log\left(1-\tfrac{1}{k}\right)-\log\left(1-\tfrac{2}{k}\right)}\ <\ x\text{.}$$ Claim 1 follows. $$\Box$$

(Be very careful about the signs involved!)

Proof of Claim 2: We need to show that $$\frac{\text{d}\alpha_{1}}{\text{d}k}\left(k\right)\ <\ 0$$ for all (real!) $$k\geq 4$$. It suffices to show that $$\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right),\ \frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)\ >\ 0\ \text{ and }\ \frac{\frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)}{\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right)}\ >\ 1$$ for such $$k$$, where we compute that $$\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right)\ =\ \log\left(2\right)\left(\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}}{1-\frac{1}{k}}\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{1}{k}\right)k\right)$$ $$\frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)\ =\ \log\left(2\right)\left(\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}}{1-\frac{2}{k}}\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{2}{k}\right)k\right)\text{.}$$ To this end, we first validate the claim that $$\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right)$$ and $$\frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)$$ are (strictly) positive for such $$k$$. Expanding via the Mercator and geometric series, \begin{align*} \frac{\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right)}{\underbrace{\log\left(2\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{1}{k}\right)k\right)}_{>\ 0}}\ &=\ \log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}}{1-\frac{1}{k}}\\ &=\ \sum_{d=1}^{\infty} -\tfrac{d^{-1}}{k^{d}}\ +\ \sum_{d=1}^{\infty} \tfrac{1}{k^{d}}\\ &=\ \sum_{d=1}^{\infty} \tfrac{1-d^{-1}}{k^{d}}\\ &>\ 0 \end{align*} and likewise \begin{align*} \frac{\frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)}{\underbrace{\log\left(2\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{2}{k}\right)k\right)}_{>\ 0}}\ &=\ \log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}}{1-\frac{2}{k}}\\ &=\ \sum_{d=1}^{\infty} -\tfrac{d^{-1}2^{d}}{k^{d}}\ +\ \sum_{d=1}^{\infty} \tfrac{2^{d}}{k^{d}}\\ &=\ \sum_{d=1}^{\infty} \tfrac{\left(1-d^{-1}\right)2^{d}}{k^{d}}\\ &>\ 0 \end{align*} (where convergence is absolute so causes no issues). Now, $$\frac{\frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)}{\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right)}\ =\ \frac{\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}}{1-\frac{2}{k}}}{\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}}{1-\frac{1}{k}}}\ \exp\left(\log\left(2\right)k\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)\text{.}$$ Let us analyze the two multiplicands separately. First, \begin{align*} \frac{\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}}{1-\frac{2}{k}}}{\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}}{1-\frac{1}{k}}}\ &=\ \frac{\sum_{d=1}^{\infty} \tfrac{\left(1-d^{-1}\right)2^{d}}{k^{d}}}{\sum_{d=1}^{\infty} \tfrac{1-d^{-1}}{k^{d}}}\\ &=\ \frac{4+8\sum_{d=3}^{\infty} \tfrac{1-d^{-1}}{k^{d-2}}+\overbrace{\sum_{d=3}^{\infty}\tfrac{\left(1-d^{-1}\right)\left(2^{d}-8\right)}{k^{d-2}}}^{>\ 0}}{1+2\sum_{d=3}^{\infty} \tfrac{1-d^{-1}}{k^{d-2}}}\\ &>\ 4 \end{align*} (for all $$k>2$$, so definitely all $$k\geq 4$$). On the other hand, \begin{align*} \frac{\log\left(\exp\left(\log\left(2\right)k\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)\right)}{\log\left(2\right)}\ &=\ k\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\\ &=\ k\left(-\sum_{d=1}^{\infty} \frac{d^{-1}2^{d}}{k^{d}}+\sum_{d=1}^{\infty} \frac{d^{-1}}{k^{d}}\right)\\ &=\ -\sum_{d=1}^{\infty} \frac{d^{-1}\left(2^{d}-1\right)}{k^{d-1}}\text{.} \end{align*} As each summand $$-\frac{d^{-1}\left(2^{d}-1\right)}{k^{d-1}}$$ is strictly increasing in $$k$$ (again, watch out for the signs), it follows that $$\frac{\log\left(\exp\left(\log\left(2\right)k\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)\right)}{\log\left(2\right)}$$ and thus $$\exp\left(\log\left(2\right)k\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)$$ itself is strictly increasing in $$k$$. As $$\exp\left(4\log\left(2\right)\left(\log\left(1-\tfrac{2}{4}\right)-\log\left(1-\tfrac{1}{4}\right)\right)\right)\ >\ \tfrac{1}{4}\text{,}$$ for $$k\geq 4$$, we conclude that \begin{align*} \frac{\frac{\text{d}\alpha_{0;2}}{\text{d}k}\left(k\right)}{\frac{\text{d}\alpha_{0;1}}{\text{d}k}\left(k\right)}\ &=\ \underbrace{\frac{\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}}{1-\frac{2}{k}}}{\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}}{1-\frac{1}{k}}}}_{>\ 4}\ \underbrace{\exp\left(\log\left(2\right)k\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)}_{>\ \tfrac{1}{4}}\\ &>\ 1\text{,} \end{align*} completing the proof of Claim 2. $$\Box$$

Proof of Claim 3: We need to show that $$\frac{\text{d}\alpha_{1}}{\text{d}k}\left(k\right)\ <\ 0$$ for all (real!) $$k\geq 4$$. It suffices to show that $$\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right),\ \frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)\ <\ 0\text{ and }\ \frac{\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)}{\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right)}\ <\ 1$$ for such $$k$$, where we compute that $$\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right)\ =\ \log\left(2\right)\left(\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}-\tfrac{\log\left(2\right)^{-1}}{k^{2}}}{1-\frac{1}{k}}\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{1}{k}\right)\left(k-\log\left(2\right)^{-1}\right)\right)$$ $$\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)\ =\ \log\left(2\right)\left(\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}-\tfrac{2\log\left(2\right)^{-1}}{k^{2}}}{1-\frac{2}{k}}\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{2}{k}\right)\left(k-\log\left(2\right)^{-1}\right)\right)\text{.}$$ To this end, we first claim that $$\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right)$$ and $$\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)$$ are (strictly) negative for such $$k$$. Expanding, \begin{align*} \frac{\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right)}{\underbrace{\log\left(2\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{1}{k}\right)\left(k-\log\left(2\right)^{-1}\right)\right)}_{>\ 0}}\ &=\ \log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}-\tfrac{\log\left(2\right)^{-1}}{k^{2}}}{1-\frac{1}{k}}\\ &=\ \sum_{d=2}^{\infty} \tfrac{\overbrace{-\log\left(2\right)^{-1}+1-d^{-1}}^{<\ 0}}{k^{d}}\\ &<\ 0 \end{align*} (for all $$k>1$$, so certainly all $$k\geq 4$$), and likewise that \begin{align*} \frac{\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)}{\underbrace{\log\left(2\right)\exp\left(\log\left(2\right)\log\left(1-\tfrac{2}{k}\right)\left(k-\log\left(2\right)^{-1}\right)\right)}_{>\ 0}}\ &=\ \log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}-\tfrac{2\log\left(2\right)^{-1}}{k^{2}}}{1-\frac{2}{k}}\\ &=\ \sum_{d=2}^{\infty} \tfrac{\left(1-0.5\log\left(2\right)^{-1}-d^{-1}\right)2^{d}}{k^{d}}\\ &=\ \tfrac{\overbrace{-2\log\left(2\right)^{-1}+2}^{<\ 0}}{k^{2}}\ +\ \tfrac{\overbrace{-4\log\left(2\right)^{-1}+\tfrac{16}{3}}^{<\ 0}}{k^{3}}\ +\ \tfrac{1}{k^{4}}\sum_{d=4}^{\infty} \tfrac{\overbrace{\left(16-8\log\left(2\right)^{-1}-16d^{-1}\right)}^{>\ 0}2^{d-4}}{k^{d-4}}\\ &<\ \tfrac{-2\log\left(2\right)^{-1}+2}{\tfrac{k^{4}}{16}}\ +\ \tfrac{-4\log\left(2\right)^{-1}+\tfrac{16}{3}}{\tfrac{k^{4}}{4}}\ +\ \tfrac{1}{k^{4}}\sum_{d=4}^{\infty} \tfrac{\left(16-8\log\left(2\right)^{-1}\right)2^{d-4}}{4^{d-4}}\\ &=\ \tfrac{\overbrace{\tfrac{256}{3}-64\log\left(2\right)^{-1}}^{<\ 0}}{k^{4}}\\ &<\ 0 \end{align*} for $$k\geq 4$$, completing the proof of the negativity of $$\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)$$ for $$k\geq 4$$. Now, $$\frac{\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)}{\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right)}\ =\ \frac{\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}-\tfrac{2\log\left(2\right)^{-1}}{k^{2}}}{1-\tfrac{2}{k}}}{\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}-\tfrac{\log\left(2\right)^{-1}}{k^{2}}}{1-\tfrac{1}{k}}}\ \exp\left(\log\left(2\right)\left(k-\log\left(2\right)^{-1}\right)\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)\text{.}$$ As before, we analyze the two multiplicands separately. First, \begin{align*} \frac{\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}-\tfrac{2\log\left(2\right)^{-1}}{k^{2}}}{1-\tfrac{2}{k}}}{\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}-\tfrac{\log\left(2\right)^{-1}}{k^{2}}}{1-\tfrac{1}{k}}}\ &=\ \frac{\sum_{d=2}^{\infty} \tfrac{\left(1-0.5\log\left(2\right)^{-1}-d^{-1}\right)2^{d}}{k^{d}}}{\sum_{d=2}^{\infty} \tfrac{-\log\left(2\right)^{-1}+1-d^{-1}}{k^{d}}}\\ &=\ \frac{\tfrac{\overbrace{2\log\left(2\right)^{-1}-2}^{>\ 0}}{k^{2}}\ +\ \tfrac{\overbrace{4\log\left(2\right)^{-1}-\tfrac{16}{3}}^{>\ 0}}{k^{3}}\ +\ \tfrac{1}{k^{4}}\sum_{d=4}^{\infty} \tfrac{\overbrace{\left(-16+8\log\left(2\right)^{-1}+16d^{-1}\right)2^{d-4}}^{<\ 0}}{k^{d-4}}}{\tfrac{\overbrace{\log\left(2\right)^{-1}-\tfrac{1}{2}}^{>\ 0}}{k^{2}}\ +\ \sum_{d=3}^{\infty} \tfrac{\overbrace{\log\left(2\right)^{-1}-1+d^{-1}}^{>\ 0}}{k^{d}}}\\ &<\ \frac{\tfrac{2\log\left(2\right)^{-1}-2}{k^{2}}\ +\ \tfrac{4\log\left(2\right)^{-1}-\tfrac{16}{3}}{4k^{2}}}{\tfrac{\log\left(2\right)^{-1}-\tfrac{1}{2}}{k^{2}}}\\ &=\ \frac{3\log\left(2\right)^{-1}-\tfrac{10}{3}}{\log\left(2\right)^{-1}-\tfrac{1}{2}}\\ &<\ 2 \end{align*} for $$k\geq 4$$. On the other hand, \begin{align*} \frac{\log\left(\exp\left(\log\left(2\right)\left(k-\log\left(2\right)^{-1}\right)\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)\right)}{\log\left(2\right)}\ &=\ -\left(k-\log\left(2\right)^{-1}\right)\sum_{d=1}^{\infty} \frac{d^{-1}\left(2^{d}-1\right)}{k^{d}}\\ &=\ -1\ +\ \sum_{d=1}^{\infty} \frac{\overbrace{-\left(d+1\right)^{-1}\left(2^{d+1}-1\right)\ +\ d^{-1}\left(2^{d}-1\right)\log\left(2\right)^{-1}}^{<\ 0}}{k^{d}}\\ &<\ -1\text{,} \end{align*} from which it immediately follows that $$\exp\left(\log\left(2\right)\left(k-\log\left(2\right)^{-1}\right)\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)\ <\ \tfrac{1}{2}\text{.}$$ We conclude that \begin{align*} \frac{\frac{\text{d}\alpha_{1;2}}{\text{d}k}\left(k\right)}{\frac{\text{d}\alpha_{1;1}}{\text{d}k}\left(k\right)}\ &=\ \underbrace{\frac{\log\left(1-\tfrac{2}{k}\right)\ +\ \frac{\tfrac{2}{k}-\tfrac{2\log\left(2\right)^{-1}}{k^{2}}}{1-\tfrac{2}{k}}}{\log\left(1-\tfrac{1}{k}\right)\ +\ \frac{\tfrac{1}{k}-\tfrac{\log\left(2\right)^{-1}}{k^{2}}}{1-\tfrac{1}{k}}}}_{<\ 2}\ \underbrace{\exp\left(\log\left(2\right)\left(k-\log\left(2\right)^{-1}\right)\left(\log\left(1-\tfrac{2}{k}\right)-\log\left(1-\tfrac{1}{k}\right)\right)\right)}_{<\ \tfrac{1}{2}}\\ &<\ 1\text{,} \end{align*} completing the proof of Claim 3. $$\Box$$

That covers our bases. Let me know if there are (inevitable) typos, I guess... $$\blacksquare$$

• trying to show $\alpha\left(k\right)\$ are strictly decreasing by derivation as k=x get complicated, i got: $\dfrac{\left(\frac{x-1}{x}\right)^{\ln\left(2\right)\,x}\,\left(\ln\left(2\right)\ln\left(\frac{x-1}{x}\right)\left(x-1\right)+\ln\left(2\right)\right)}{x-1}-\dfrac{\left(\frac{x-2}{x}\right)^{\ln\left(2\right)\,x}\,\left(\ln\left(2\right)\ln\left(\frac{x-2}{x}\right)\left(x-2\right)+2\ln\left(2\right)\right)}{x-2}$ Commented Aug 6, 2023 at 22:45
• @mathxxx Honestly, I was really hoping that someone else would take my suggestion and fill in the (very messy) details. Well, no one has, so I've written up the extent of the argument that I had in mind.
– Rafi
Commented Aug 10, 2023 at 22:54