I believe that I am following Hanson's proof up until this point (page 36):

$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}$$

where $e$ is Euler's constant, $a_1=2$, $a_{i+1} = a_1 a_2 \dots a_i + 1$ so that:

  • $a_2 = 3$
  • $a_3 = 7$

$k$ is the value where $a_k \le n < a_{k+1}$

$w^n$ is not important for my question but it is defined as $w = a_1^{1/a_1}a_2^{1/a_2}\dots$

I am unclear at this step where:

$$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}< e^{k - 3/2}n^{k - 3/2}w^n$$

where $k > 2$ since $n < a_1 a_2 \dots a_k$

Here's what I understand:

  • Since each $a_i > 1$, I am clear why:

$$\dfrac{(en)^{k-1+1/(a_{k+1}+1)}w^n}{(a_1)^{(a_1-1)/a_1}(a_2)^{(a_2-1)/a_2}\dots(a_m)^{(a_k-1)/a_k}}< e^{k - 1 + 1/(a_{k+1}+1)}n^{k - 1 + 1/(a_{k+1}+1)}w^n$$

  • If he meant to use $k - \frac{3}{4}$, then I would be clear since:

$$(en)^{k-1+1/(a_{k+1}+1)}w^n < e^{k - 3/4}n^{k - 3/4}w^n$$

  • If $k - \frac{3}{2}$ is correct, then I would appreciate help in understanding how this follows.

It is not obvious to me that this is true or why it is true.

  • $\begingroup$ @Clayton, I explored that direction but couldn't see how $a_i$ would relate to $(en)$. I think that you are right that this is where the answer should be. If I figure it out, I will update my question to show what I came up with. $\endgroup$ – Larry Freeman Oct 11 '18 at 18:58
  • $\begingroup$ Only a partial answer: the denominator simplifies to $(a_1\cdots a_k)(a_1^{-1/a_1}\cdots a_k^{-1/a_k})$. Using that $n\leq a_1\cdots a_k$ (this is by hypothesis), we can say that $(\text{right-hand side})\leq\frac{1}{n(a_1^{-1/a_1}\cdots a_k^{-1/a_k})}$. Now part of the problem that remains is the product of $a_i^{1/a_i}$ tends to a limit that is greater than $e$. With a negative exponent (in the denominator), this seems to imply that the exponent of $e$ should increase by a quantity of at least $1$. $\endgroup$ – Clayton Oct 11 '18 at 19:35
  • $\begingroup$ @Clayton, I think that this suggests that it may be a mistake. Hanson shows that $a_1 a_2 \dots a_k > n$ and $a_1^{1/a_1}a_2^{1/a_2}\dots < 2.952$ so that the denominator is roughly $n/2.952$ which is smaller than $n/e$. $\endgroup$ – Larry Freeman Oct 12 '18 at 0:35
  • $\begingroup$ Larry, that was precisely my point. @mathlove caught the mistake, though! This makes it all alright :) $\endgroup$ – Clayton Oct 12 '18 at 15:54

First of all, from your previous question, I think (2.6) should be $$C(n) < \dfrac{(en)^{k-1+1/(a_{k+1}\color{red}{-}1)}w^n}{a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}}\lt e^{k - 3/2}n^{k - 3/2}w^n$$ (see the minus sign in red in the numerator)

To prove $$\dfrac{(en)^{k-1+1/(a_{k+1}-1)}w^n}{a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}}\lt e^{k - 3/2}n^{k - 3/2}w^n\tag1$$ it is sufficient to prove $$(en)^{k-1+1/(a_{k+1}-1)-(k-3/2)}\lt a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k},$$ i.e. $$(en)^{1/2+1/(a_{k+1}-1)}\lt a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k}$$ Since $n\le a_1a_2\cdots a_k$, it is sufficient to prove $$(ea_1a_2\cdots a_k)^{1/2+1/(a_{k+1}-1)}\lt a_1^{(a_1-1)/a_1}a_2^{(a_2-1)/a_2}\dots a_k^{(a_k-1)/a_k},$$ i.e. $$e^{1/2+1/(a_{k+1}-1)}\lt a_1^{1/2-1/a_1-1/(a_{k+1}-1)}a_2^{1/2-1/a_2-1/(a_{k+1}-1)}\cdots a_k^{1/2-1/a_k-1/(a_{k+1}-1)},$$ i.e. $$e^{1/2+1/(a_{k+1}-1)}2^{1/(a_{k+1}-1)}\lt a_2^{1/2-1/a_2-1/(a_{k+1}-1)}\cdots a_k^{1/2-1/a_k-1/(a_{k+1}-1)}$$ Since $a_i^{1/2-1/a_i-1/(a_{k+1}-1)}\gt 1$ for $2\le i\le k-1$, It is sufficient to prove $$e^{1/2+1/(a_{k+1}-1)}2^{1/(a_{k+1}-1)}\lt a_k^{1/2-1/a_k-1/(a_{k+1}-1)}\tag2$$ for $k\ge 3$.

Note here that seeing $(2)$ as an inequality on $k$, we see that the LHS of $(2)$ is decreasing and that the RHS of $(2)$ is increasing, so it is sufficient to prove $(2)$ for $k=3$, which is equivalent to $$2e^{22}\lt 7^{14}$$ which is indeed true.

Therefore, $(1)$ is true.

  • $\begingroup$ Wow. Great job. I was convinced that Hanson had made another mistake. I am very glad to hear that I was wrong. $\endgroup$ – Larry Freeman Oct 12 '18 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.