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Reading through a paper and it is stated that $k(1-1/d)^{k-1} < 1$ implies that $k < (1+o(1))d \log d$ without proof (assume $k,d > 1$). How is this fact derived? I assume that the $d \log d$ term comes from $\log(1-1/d) < \log \log d$ but apart from that it is not obvious to me.

Doc Stories
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  • I suggest instead considering the intuition $1-1/d\approx e^{-1/d}$ and taking logarithms of both sides of the assumed inequality. – Greg Martin Oct 24 '23 at 19:33
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    $k(1-1/d)^{k-1} > 1$ or $k(1-1/d)^{k-1} < 1$? – River Li Oct 24 '23 at 23:52
  • @RiverLi the theorem assumption is < 1 (see Thm 11 in the paper) – Doc Stories Oct 25 '23 at 01:18
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    @DocStories I think that you misunderstood theorem 11. It should be $k(1 - 1/d)^{k-1} < 1$ implies $k > R(d)$. Note that $\lim_{k\to \infty} k(1-1/d)^{k-1} = 0 < 1$. – River Li Oct 25 '23 at 01:33
  • @RiverLi ah, I see. So the assumption implies that $k > R(d)$ but it's still not clear that this further implies $R(d) \le (1 + o(1))d \log d$? – Doc Stories Oct 25 '23 at 02:04
  • It needs to be checked, by I guess it works as follows. Since $1+x\le e^x$ for each real $x$, it suffices to choose $k$ such that $ke^{-(k-1)/d}<1$. If $k=(1+o(1))d\log d$ then $ke^{-(k-1)/d}=(1+o(1))(\log d)d^{-o(1)}$. – Alex Ravsky Oct 25 '23 at 03:01
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    @DocStories The first part of Theorem 11 proved $R(d) < k$ for any $k> 1$ satisfying $k(1 - 1/d)^{k-1} < 1$. For fixed $d > 1$, let $f(d) := \inf {k \in \mathbb{R}_{> 1} : k(1 - 1/d)^{k-1} < 1}$. Then we have $R(d)\le f(d)$. We can prove that $f(d) < (1 + o(1))d\ln d$. This completes the proof. – River Li Oct 25 '23 at 06:45
  • @RiverLi ah! I see the logic now. Thank you. – Doc Stories Oct 25 '23 at 13:15
  • @DocStories You are welcome. – River Li Oct 25 '23 at 13:18
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    According to the discussion, the problem can be rephrased as follows. Let $a \in \mathbb{R}$. Let $d > 1$. If $a < k$ for all $k>1$ satisfying $k(1 - 1/d)^{k-1} < 1$, then $a < (1 + o(1))d\ln d$. – River Li Oct 31 '23 at 03:48

1 Answers1

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According to the comments, the problem can be rephrased as follows.

Problem. Let $a \in \mathbb{R}$. Let $d > 1$. If $a < k$ for all $k>1$ satisfying $k(1 - 1/d)^{k-1} < 1$, then $a < (1 + o(1))d\ln d$.

Proof. Consider $d > 4$. Choose $$k = \left(1 + \frac{\ln(\ln d) + \frac{1}{d}}{\ln d - 1}\right)d\ln d.$$

We claim that $k(1 - 1/d)^{k - 1} < 1$. Indeed, we have \begin{align*} \ln k + (k - 1)\ln(1 - 1/d) &= \ln (d\ln d) + \ln\frac{k}{d\ln d} + (k - 1)\ln(1 - 1/d)\\[6pt] &< \ln (d\ln d) + \frac{k}{d\ln d} - 1 + (k - 1)(-1/d)\\[6pt] &= 0 \end{align*} where we use $\ln x \le x - 1$ for all $x > 0$, and $\ln (1 - 1/d) < -1/d$.

Thus, we have $$a < \left(1 + \frac{\ln(\ln d) + \frac{1}{d}}{\ln d - 1}\right)d\ln d = (1 + o(1))d\ln d.$$

We are done.

River Li
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  • Already upvote. Nice solution! Btw, could you please think of problem (https://math.stackexchange.com/questions/4796969/finding-small-max-limits-abbcca-1-dfrac-sqrtabc-sqrtbca-sqrtca?noredirect=1#comment10203442_4796969) help me? Thank you –  Nov 01 '23 at 04:20