# Two (strictly related) proofs by induction of inequalities.

This is a question I originally asked on MSE, receiving no answer, even with a bounty (which expired) on it. Therefore I am crosslinking in order to prevent duplication of effort: see here for the original question.

Predictably, I am stuck with the inductive steps. Let $a_n=\prod_{i=1}^m p_i^{b_i}$, where $b_1=9;b_2=5;b_3,b_4=3;b_5,b_6=2;b_7,\ldots ,b_{m}=1$, $p_i$ is the $i$-th prime and set $\lim_{n\to \infty}\frac{\log a_n}{p_m}=1$. Suppose also this ratio converges to $1$ faster than $\displaystyle\frac{p_{m(n)+1}}{p_{m(n)}}$, so that if $n$ is large enough, we always have $\log a_n<p_{m+1}$.

I want to prove that for sufficiently large $n$, with $c$ being a constant and $q(n)<m$, if $$\frac{c}{\log \log a_n}<\frac{\left(1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}\right)^m}{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)}, \tag{1}$$ then the following statements are true:

$$\frac{c}{\log \left(\log a_n+\log p_q\right)}<\\\frac{\left(1+\prod_{i=1}^{q-1}\left(p_i^{b_i+1}-1\right)^{1/m}\cdot\left(p_q^{b_q+2}-1\right)^{1/m}\cdot\prod_{i=q+1}^{m}\left(p_i^{b_i+1}-1\right)^{1/m}\right)^m}{\prod_{i=1}^{q-1}\left(p_i^{b_i+1}-1\right)\cdot\left(p_q^{b_q+2}-1\right)\cdot\prod_{i=q+1}^{m}\left(p_i^{b_i+1}-1\right)}; \tag{2}$$

$$\frac{p_{m(n)+1}}{p_{m(n)+1}-1}\frac{c}{\log \left(\log a_n+\log p_{m(n)+1}\right)}<\frac{\left(1+\prod_{i=1}^{m(n)+1}\left(p_i^{b_i+1}-1\right)^{1/(m(n)+1)}\right)^{m(n)+1}}{\prod_{i=1}^{m(n)+1}\left(p_i^{b_i+1}-1\right)}. \tag{3}$$

To clear it up, in $(2)$ we have $a_n\cdot p_q=a_{n+1}$, in $(3)$ instead $a_n\cdot p_{m(n)+1}=a_{n+1}$. Namely, we're considering two different cases of how the sequence $a_n$ evolves: in $(3)$ the $n+1$-th term is given by the product of the $n-$th term times the $n+1$-th prime; in $(2)$ the $n$-th term is multiplied by a prime less than the $n+1$-th.

$(2)$ is fairly intuitive, as the LHS goes to $0$ as $n\to \infty$ while the RHS goes to $1$, but that doesn't tell me so much since if the former is larger than $1$ and slightly smaller than the latter, I cannot say a priori that the LHS is sufficiently fast in its convergence to $0$, to be always less than the RHS. On the other hand, it is only my istinct that says $(3)$ holds, but I might be wrong.

Here is how I tackled both inequalities, hoping to simplify things a bit (and not "too much"). Call $L_t$ and $R_t$ respectively the LHS and RHS of $(1)$, $(2)$ and $(3)$. So $(2)$ is the same as $$L_1 \frac{L_2}{L_1}<R_1\frac{R_2}{R_1},$$ and since $L_1<R_1$ by hypothesis, $(2)$ is implied by $$\frac{\log \log a_n}{\log \left(\log a_n+\log p_q\right)}<\\ \frac{p_q^{b_q+1}-1}{p_q^{b_q+2}-1}\left(\frac{1+\prod_{i=1}^{q-1}\left(p_i^{b_i+1}-1\right)^{1/m}\cdot\left(p_q^{b_q+2}-1\right)^{1/m}\cdot\prod_{i=q+1}^{m}\left(p_i^{b_i+1}-1\right)^{1/m}}{1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}}\right)^m.\tag{4}$$ Similarly, $(3)$ follows from $$\frac{p_{m+1}}{p_{m+1}-1}\frac{\log \log a_n}{\log \left(\log a_n+\log p_{m+1}\right)}<\\ \left(1+\prod_{i=1}^{m+1}\left(p_i^{b_i+1}-1\right)^{1/(m+1)}\right)\left(\frac{1+\prod_{i=1}^{m+1}\left(p_i^{b_i+1}-1\right)^{1/(m+1)}}{1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}}\right)^m.\tag{5}$$ This said, I do not know how to prove $(4)$ and $(5)$ either. Any ideas? Thanks in advance.

EDIT: I have succeeded in making sufficient to prove the inequalities having $p_{m+1}-1$ instead of $p_m$, which somewhat might be a tiny bit easier.

• Is $a_n=\prod\limits_{k=1}^mp_i^{b_i}$ ? – robjohn Dec 12 '14 at 15:55
• @robjohn Yes, exactly. – Vincenzo Oliva Dec 12 '14 at 16:02
• What is the motivation behind your question? – Jose Arnaldo Bebita-Dris Dec 31 '14 at 14:59
• @Jose This is something I need for a bigger thing, in a future paper. – Vincenzo Oliva Dec 31 '14 at 15:15
• I've clarified something and fixed a typo. – Vincenzo Oliva Jan 1 '15 at 13:22