# 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 $\log a_n<p_{m(n)+1}$.

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 $$\frac{p_{m(n)+1}}{p_{m(n)+1}-1}\frac{\log \log a_n}{\log \left(\log a_n+\log p_{m(n)+1}\right)}<\\ \left(1+\prod_{i=1}^{m(n)+1}\left(p_i^{b_i+1}-1\right)^{1/(m(n)+1)}\right)\left(\frac{1+\prod_{i=1}^{m(n)+1}\left(p_i^{b_i+1}-1\right)^{1/(m(n)+1)}}{1+{\prod_{i=1}^m\left(p_i^{b_i+1}-1\right)^{1/m}}}\right)^m.\tag{5}$$ 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 $p_{m(n)+1}-1$ instead of $p_m$, which somewhat might be a tiny bit easier.

• Let's get the notation straight before talking about anything else: the opening phrase "Let $p_m$ be the largest prime factor of $a_n$" puts me in such a beautiful state of blissful ignorance about the indexation in the following three story formulae, that the proverbial ram looking at the new gate is an example of perfect comprehension and deep understanding compared to me looking at them ;-) Dec 27 '14 at 16:50
• @fedja I'm sorry for that. Though I'm not totally sure how I could improve the exposition. Does stating $a_n=\prod_{i=1}^m p_i^{b_i}$ help? Anyway, perhaps you might be interested in this question, whose inequality implies $(1)$ here, and I guess looks a bit nicer (I hope I'll manage to keep your attention on my problem :D) Dec 27 '14 at 17:22
• Now it looks a bit better :-). However I'm still a bit perplexed by the notation $a_n$ and the phrase $n\to\infty$ because there is no running $n$ anywhere else. Also, what on Earth is $p_{m+1}$ in this case? Dec 27 '14 at 23:22
• @fedja As for $\lim_{n→∞}\frac{\log a_n}{p_m}$, I considered $p_m$ a function of $n$, as it is the largest prime factor of $a_n$. But I guess I can say $n,m→∞$. $p_{m+1}$ is the prime that follows $p_m$. So for example if $a_n=2^{10}⋅3^6⋅5^4, p_{m+1}=7$. Dec 28 '14 at 1:03
• The definition of $a_n$ suggests that $m$ depends on $n$, but then the $\lim_{n,m\to\infty}$ suggests that $m$ does not depend on $n$, and that $(n,m)$ is just a pair of non-negative independent integers. Was it $a_{n,m}$ instead of $a_n$? Not clear. Dec 31 '14 at 23:31