About the recursive inequality $w_p \geq (1-\frac {\pi}n)w_{p-2n} + 2\pi + o(1)$

Question

Suppose we have a non-decreasing sequence of positive real numbers that tend to infinity: $0<w_1\leq w_2\leq w_3\leq...$ It is known that:

• For every $n$ and $p\geq 2n$, we have $w_p \geq (1-\frac {\pi}n)w_{p-2n} + 2\pi + o(1)$, where the $o(1)$ is with respect to $n$.
• The sequence satisfies $\lim_n \frac{w_n}{\sqrt n}=a$, for some constant $0<a\leq \pi$.

The question is, from these facts, is it possible to deduce $a=\pi$?

By playing with the above recursive inequality and an inductive argument, I managed to get some non-trivial lower bound on $a$, but was not sure how to push the lower bound to $\pi$.

Thank you.

Answer 1

I will show that $a \ge \sqrt{\pi}$ which I assume is what you wanted to ask (and it complements counterexample of Iosif Pinelis which appeared while I was writing this answer, showing that this is sharp).

Let us rewrite your inequality as $$w_p - w_{p-2n} \ge 2\pi + o(1) - \frac{\pi w_{p-2n}}{n}.$$

Then first idea is that we want to pick some sequence of (say, even) numbers $p_k$ and, denoting $n_k = \frac{p_k - p_{k-1}}{2}$, apply the inequality to $p = p_k$ and $n = n_k$ so the left-hands side becomes $w_{p_k} - w_{p_{k-1}}$ and then telescope. The game now becomes of choosing an appropriate sequence of $p_k$ (or $n_k$, since they can be directly written through each other).

My analytical intuition tells me that we should have $\frac{w_{p-2n}}{n}$ be bounded away from $0$ and from $\infty$ (if it is too big then the inequality is useless, and if it is too small then basically we will only have $2\pi$ which also doesn't seem to be too powerful on its own). If $n$ is comparable to $p$ then $\frac{w_{p-2n}}{n} \lesssim \frac{\sqrt{p}}{n} = o(1)$, so this is no good. So, $n = o(p)$ and hence $w_{p-2n}\asymp a\sqrt{p}$. Thus, we should take $n \sim \sqrt{p}$. An example of such sequence is $p_k = 2\left[\gamma k^2\right]$ ($2$ is only to make it even). Then $n_k \asymp 2\gamma k$ (in particular, it does tend to $\infty$ so $o(1)$ is $o(1)$ in our case). Plugging this in we get $$w_{p_k} - w_{p_{k-1}} \ge 2\pi + o(1) - \frac{\pi a \sqrt{p_k - 2 n_k}}{n_k}= 2\pi + o(1) - \frac{\pi a\sqrt{2\gamma k^2}}{2\gamma k} = 2\pi + o(1) - \frac{\pi a}{\sqrt{2\gamma}},$$

where I will leave it to you to check that all the errors introduced by our approximations can be absorbed by $o(1)$.

Telescoping, we get

$$w_{p_k} - w_{p_0} \ge k\left(2\pi - \frac{\pi a}{\sqrt{2\gamma}}\right).$$

The number $w_{p_0}$ is just a constant, so it can be discarded as well, and $w_{p_k}$ is approximately $a\sqrt{p_k}\asymp k a\sqrt{2\gamma}$. Dividing by $k$ we get

$$a\sqrt{2\gamma} \ge 2\pi - \frac{\pi a}{\sqrt{2\gamma}},$$ and now we just optimize in $\gamma$. I will spare you looking at me taking the derivative, picking $\gamma =\frac{\pi}{2}$ we get $a\ge \sqrt{\pi}$.

Answer 2

No. E.g., if $w_n=\sqrt\pi\,\sqrt n$, then for all natural $n\ge4$ and all natural $p\ge2n$ $$w_p-\Big(1-\frac {\pi}n\big)w_{p-2n}\ge\sqrt{4-2\frac\pi n} \,\pi=2\pi + o(1),$$ so that your recursive inequality holds.