Proving $\frac{p_{n+1}^2-p_{n+1}^{-C}}{p_{n+1}^2-1}>\frac{f(n) p_n \log p_n }{p_{n+1} \log p_{n+1}},$ where $f(n)\to1$, for some constant $C>1$

Question

Are there two constants $C_1$, $C_2>1$ such that for large enough $n$ $$\frac{p_{n+1}^2-p_{n+1}^{-C_1}}{p_{n+1}^2-1}>\frac{2 C_2 p_{n+1} \log p_{n+1}-1}{2 C_2 p_{n} \log p_{n}-1} \left(\frac{p_n \log p_n}{p_{n+1} \log p_{n+1}}\right)^n \ \prod_{k=2}^{n-1} \frac{p_k^2C_2p_{n+1}\log p_{n+1}-1}{p_k^2C_2p_{n}\log p_{n}-1} \ \ \ \ ?$$

After other attempts (failures), here's what I did to show the thruthfulness of the statement:

Let $m=p_{n+1}$ and set $$\ g(m)>\frac{2 C_2 p_{n+1} \log p_{n+1}-1}{2 C_2 p_{n} \log p_{n}-1} \left(\frac{p_n \log p_n}{p_{n+1} \log p_{n+1}}\right)^{n-1} \ \prod_{k=2}^{n-1} \frac{p_k^2C_2p_{n+1}\log p_{n+1}-1}{p_k^2C_2p_{n}\log p_{n}-1} $$ to write the stronger inequality $$\frac{m^2-m^{-C_1}}{m^2-1}>\frac{g(m) (m-2) \log (m-2)}{m \log m},$$ which, since the LHS converges to $1$ strictly from above and, provided that $g(m)\to 1$, clearly the RHS goes to $1$ as well, is implied by $$D\left(\frac{g(m) (m-2) \log (m-2)}{m \log m}\right)>0.$$ This reduces to $$\frac{g'(m)}{g(m)}>\frac{1}{m} - \frac{1}{m-2} - \frac{1}{(m-2)\log (m-2)}.$$ However I can't see what to do with this to prove the original inequality. How does one proceed? Or how does one prove it with a different approach? Thank you.

Answer 1

I have had a go at trying to prove it by simply estimating all the different terms in the equation to find the asymptotic behavior of two sides of the equation. Below $\sim$ will denote an asymptotic value.

Using $p_n\sim n\log n$ we get the following estimates

$$A_n \equiv \frac{p_n\log p_n}{p_{n+1}\log p_{n+1}} \sim 1 - \frac{1}{n}$$

$$B_n \equiv \frac{1}{p_n\log p_n}-\frac{1}{p_{n+1}\log p_{n+1}} \sim \frac{1}{n^2\log^2(n)}$$

Using this we find

$$\frac{2 C_2 p_{n+1} \log p_{n+1}-1}{2 C_2 p_{n} \log p_{n}-1} \sim 1+\frac{B_n}{2C_2} \sim 1+\frac{1}{2C_2n^2\log^2(n)}$$

$$\left(\frac{p_{n}\log p_{n}}{p_{n+1}\log p_{n+1}}\right)^{n-2}\times \prod_{k=2}^{n-1}\frac{C_2p_k^2p_{n+1}\log p_{n+1}-1}{C_2p_k^2p_{n}\log p_{n}-1} = \\\prod_{k=2}^{n-1}\left[1 + \frac{B_n}{C_2 p_k^2} + \mathcal{O}\left(\frac{1}{p_k^4n^2\log^4 n}\right)\right]\sim 1 + \frac{1}{C_2n^2\log^2 n}\sum_{k=2}^{n-1}\frac{1}{p_k^2} + \mathcal{O}\left(\frac{1}{n^2\log^4 n}\right)$$

and since $\sum_{k=2}^\infty \frac{1}{p_k^2} < \frac{\pi^2}{6}$ the prefactor to the second term is bounded by a $\mathcal{O}(1)$ constant.

Combinding the results above gives us

$$LHS = 1 + \frac{1-p_{n+1}^{-C_1}}{p_{n+1}^2-1} \sim 1 + \frac{1}{n^2\log^2(n)}$$

for $C_1>0$ and

$$RHS \sim \left(1+\frac{1}{2C_2n^2\log^2(n)}\right)\left(1-\frac{1}{n}\right)^2\left(1 + \frac{\mathcal{O}(1)}{C_2n^2\log^2 n}\right) \sim 1 - \frac{2}{n} + \frac{\mathcal{O}(1)}{C_2 n^2\log^2(n)}$$

and it follows that LHS$-$RHS $ \sim \frac{2}{n} > 0$ for large enough $n$ independent of $C_1,C_2$ (as long as $C_1>0$).

If the $(A_n)^n$ term on the RHS is replaced by $(A_n)^{n-2}$ (removing the middle term in the estimate above) it becomes more interesting as the two sides have comparable convergence to $1$, but by choosing $C_2$ large enough it still follows that LHS$ > $RHS for large enough $n$.

Answer 2

Probably I'm wrong but I hope that this answer could help you.
In this problem I interpreted:
$$1)-p_n\ is\ the\ n-th\ prime\ number$$ $$2)-f(n)\ is\ a\ generic\ sequence\ which\ converges\ to\ 1\ for\ n\ large\ enough$$ $$3)-the\ proposition\ must\ be\ true\ for\ n\ large\ enough$$ Said that,prove that: $$\frac{p_{n+1}^2 - p_{n+1}^{-C}}{p_{n+1}^2 - 1} > f(n)\ \frac{p_{n}\ ln(p_{n})}{p_{n+1}\ ln(p_{n+1})}$$ It's equivalent to prove: $$\frac{p_{n+1}^2 - p_{n+1}^{-C}}{p_{n+1}^2 - 1}\ \frac{p_{n+1}\ ln(p_{n+1})}{p_{n}\ ln(p_{n})} >\ f(n)$$ let's suppose: $$s(n)\ :=\ \frac{p_{n+1}^2 - p_{n+1}^{-C}}{p_{n+1}^2 - 1}\ \frac{p_{n+1}\ ln(p_{n+1})}{p_{n}\ ln(p_{n})}$$ Then we have:$$s(n) > f(n)$$ It's not difficult to prove that: $$\lim_{n \to \infty}s(n)\ =\ 1\ for\ each\ c>1$$ Infact the Prime number theorem implies that: $$p_n\ \sim\ n\ ln(n)$$ But if we suppose: $$f(n)\ :=\ e^{s(n) - 1}$$ We can verify that: $$\lim_{n \to \infty}f(n)\ =\ 1 $$ $$ s(n)\ <\ f(n)\ =\ e^{s(n)\ -\ 1}$$ Because: $$ x \le e^{x\ -\ 1}\ with\ x\in\mathbb{R} $$ That should be a contraddiction.
I'm sorry for the form but it's my first answer.