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Let $p_{n}$ denotes the $n$-th prime number. Is it true that $p_{n}+p_{n+1}>p_{n+2}$ for all $n\geq 2\ ?$

Bart Michels
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Sergiy Kozerenko
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1 Answers1

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Yes. Check $n=2$ and $n=3$ by hand. A variant of Bertrand's postulate assures, that there are two primes between $p_n$ and $2p_n$, when $n\geq 4$. Let's call the bigger one $p_k$, then $k\geq n$+2 and $$p_n+p_{n+1}\geq 2p_n > p_k \geq p_{n+2}$$

Community
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Tomas
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  • +1 It is more of a generalization, than a variant. For all $\alpha > 1$, there exists an $n$ such that for $m > n$, there is a prime in the interval $m, \alpha m$. The 'variant' that you are using is $\alpha = \frac{4}{3}$, with $n=12$. – Calvin Lin Oct 18 '13 at 14:57
  • Thanks, @CalvinLin, I agree with the generalization and $\frac{4}{3}$ of course works, since this yields two primes between $m$ and $\frac{16}{9}m<2m$. But why did you pick precisely $\frac 43$? – Tomas Oct 18 '13 at 17:32
  • It is one of those well-known facts that are quoted without proof in math competitions. – Calvin Lin Oct 18 '13 at 18:07