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Let $p_n$ denote the $n$-th prime. Let $k>0$ be an integer.

Assume that there exists $N_0$ s.t. $p^{k}_{N_{0}+1} < p_1 p_2 ... p_{N_{0}}$. Show that $$p^{k}_{n+1}< p_1 p_2 ... p_{n}, \space \forall n \geq N_0 $$

Then the question says, you can assume Bertrand's postulate.

EDIT:

$p^{k}_{N_{0}+1} < p_1 p_2 ... p_{N_{0}} $ is our base case and holds by question.

Assume it holds for all $ n \geq N_0$ WTS holds for $n+1$. Since it holds for all $n$ s.t $ n \geq N_0$, I think we can assume that $n \geq \max \{ N_0 , 2^k \}$ then prove it for $n+1$

We know by Bertrand's postulate that $$ p_{n}< p_{n+1} <2p_{n} \Rightarrow p^{k}_{n}< p^{k}_{n+1} <2^{k}p^{k}_{n} $$

but by IH $$2^{k}p^{k}_{n}<2^k p_1...p_{n-1} \Rightarrow p^{k}_{n+1}<2^k p_1...p_{n-1}$$

$ 2^k \leq n \leq p_n $ hence $p^{k}_{n+1}< p_1...p_{n-1}p_{n}$ as desired.

rtybase
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Faust
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  • Why can you assume that $n \ge \max(N_0,2^k)$ ? For instance, with $k=3$, $n=N_0=5$, you don't have $5 \ge \max(5,8)$. I think that for the IH you could try to prove first that $\frac{p_{n+1}^3}{p_n^4} \le 1$ for $n \ge 3$. So if $N_0 \ge 3$ it's ok, and if $N_0 \le 2$, well, $k=1$ so it's also ok. I don't know if it is hard to prove the inequality that I suggested (but at least it seems true on small values) – charmd Jan 14 '18 at 08:17
  • What im trying to do is use induction to get around the thing your stating, By induction i assume that it holds for all $N_0 \leq \ell \leq 2^k $ then i want to prove it for $\ell +1 $ induction assumes it hold for an arbitrary number of cases. i have never seen induction used in such a way but it doesn't seem unreasonable? – Faust Jan 14 '18 at 16:12

1 Answers1

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Observation there is one example $k=2$, according to Bonses's inequality $$p_1p_2...p_n > p_{n+1}^2$$ for $\forall n\geq 4= 2^2$. This $2^2=2^k$ is the key.

For $k=3$ we can see that $2\cdot3\cdot5\cdot7\cdot11 > 13^3$, but we can extend it to $N_0=2^3$ $$2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19 > 23^3$$ and we have the initial case for induction, using the technique you suggested (which looks good by the way).

For $k=4$, $N_0=2^4$ $$2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43\cdot47\cdot53 > 59^4 $$

Then the problem is to find the initial case for each $k>4$.

Now that we have established that the problem is in finding the initial case let's use induction for this problem. We already established that we have initial cases for $k \in \{2,3,4\}$, let's assume there is one for $N_0=2^k$ and prove that this leads to finding an initial case for $N_0=2^{k+1}$ $$\prod\limits_{t=1}^{2^k} p_t > p_{2^k+1}^{k} \Rightarrow \prod\limits_{t=1}^{2^k+1} p_t > p_{2^k+1}^{k+1} \Rightarrow \\ \left(\prod\limits_{t=1}^{2^k+1} p_t\right)\left(\prod\limits_{t=2^k+2}^{2^{k+1}} p_t\right) > p_{2^k+1}^{k+1}\left(\prod\limits_{t=2^k+2}^{2^{k+1}} p_t\right) \Rightarrow \\ \left(\prod\limits_{t=1}^{2^{k+1}} p_t\right) > p_{2^k+1}^{k+1}\left(\prod\limits_{t=2^k+2}^{2^{k+1}} p_t\right) \overset{p_m>m}{>} (2^k+1)^{k+1} \left(\prod\limits_{t=2^k+2}^{2^{k+1}} t\right)=\\ (2^k+1)^{k+1} (2^k+2)(2^k+3)(2^k+4)...2^{k+1}> (2^k+1)^{k+1} (2^k+1)^{2^k-1}> ...$$ It is worth mentioning that $2^k-1\geq 3k+3, \forall k\geq 4$ $$...>(2^k+1)^{k+1} (2^k+1)^{k+1} (2^k+1)^{k+1} (2^k+1)^{k+1}=(2^k+1)^{4(k+1)}>...$$ and $(2^k+1)^2 >2^{k+1}+1 $ $$...>(2^{k+1}+1)^{2(k+1)}$$ As a result we have $$\prod\limits_{t=1}^{2^{k+1}} p_t > (2^{k+1}+1)^{2(k+1)} \tag{1}$$

It is worth mentioning that $$\sqrt{p_m} < m \Rightarrow p_m < m^2$$ so $$p_{2^{k+1}+1}< (2^{k+1}+1)^2 \tag{2}$$ Now, combining $(1)$ and $(2)$ $$\prod\limits_{t=1}^{2^{k+1}} p_t > (2^{k+1}+1)^{2(k+1)} > p_{2^{k+1}+1}^{k+1} $$ So, there is always an initial case.

rtybase
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  • This is an excellent post and honestly i am quite surprised that you were able to show that the $N_0$ exist without using the prime number theorem! I am still unsure though if in the question how i used induction in an odd way is valid argument i mean it seems reasonable... – Faust Jan 14 '18 at 16:24
  • I spent nearly 2 good hours on spotting the pattern from a few examples :) ... the good thing is that you don't need $n \geq \max { N_0 , 2^k }$ anymore, you can always start with $N_0=2^k$. Everything else looks valid. – rtybase Jan 14 '18 at 16:28