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.
Best Answer
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.