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prove that $p_{n+1}\leq(2^{2^n}+1)$.where $p_{n+1}$ is $(n+1)^\text{th}$ prime.

i am doing it using induction. for $n=1$, it is true. let true for $n.$

now we want to show that $p_{n+2}\leq{2^{2^{n+1}}}+1.$ here i am using the fact that for given natural number $n$, there exist a prime between $n$ and $2n$. using this result for $n=p_{n+1}$, we get $p_{n+2}\leq2p_{n+1.}$

now using induction our result follow.

is there any more elementary method to prove this, mean without using that result from number theory

Michael Hardy
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Eklavya
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2 Answers2

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The proof will be by strong induction.

Notice it hold for $n=0$ since $p_1=2\leq 2^{2^0}+1$

Now notice $p_{n+1}\leq p_1p_2\dots p_n+1\leq \color{red}{(2^{2^{0}}+1)(2^{2^{1}}+1)\cdots (2^{2^{n-1}}+1)}+1$

When we expand this product we get $\color{red}{1+2+2^2+\cdots + 2^{2^{2^{n}-1}}}+1=\color{red}{2^{2^n}-1}+1$

Michael Hardy
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Asinomás
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  • can you please explain why when we expand product we get this GP? – Eklavya Jul 30 '16 at 15:56
  • I think of it in binary, there are $2^n$ summands when you expand that product, and each one is a different number in binary. The idea is that every factor represents a "bit". I hope it helps, you can also prove it by induction. – Asinomás Jul 30 '16 at 16:00
  • Which combination give you the summand $2^{17}$?? notice $17=2^{2^4}+2^1$. So the summand $2^{17} $ comes from $2^{2^0}\times 1 \times 1 \times 2^{2^4}\times 1 \times 1\times 1 \dots $ – Asinomás Jul 30 '16 at 16:03
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Use the Bertrand's Postulate and the problem is trivial by induction. As for $n=1$ the claim is true, assume it holds for some $k \in \mathbb{N}$. Then using the weaker form of the Postulate we have that there exist a prime between $2^{2^k}-1$ and $2(2^{2^k}-1) - 1$. Obviously we have that $2(2^{2^k}-1) - 1 = 2^{2^k + 1} - 1 \le 2^{2^{k+1}} - 1$. So we have that $p_{k+2} \le q \le 2^{2^{k+1}} - 1$. Hence the proof.

NOTE: $q$ is the prime, whose existence is guaranteed by Bertrand's Postulate, which is either equal to $p_{k+2}$ or greater than it.

Stefan4024
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