Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?

up vote 9 down vote accepted

I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.

By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $\varepsilon >\frac{1}{5}$, there exists an $n_{\varepsilon}\in \mathbb{N}$ such that for all $n\geq n_{\varepsilon}$,


In particular, taking $\varepsilon = \frac{1}{4}$, this implies that, for every sufficiently large $k \in \mathbb{N}$, we have that

$$p_{k+2} \leq (1.25)^{2}p_{k} < 2p_{k}$$

and whence, for every sufficiently large $k \in \mathbb{N}$, it is certainly the case that

$$p_{k+2} < 2p_{k} < p_{k}+p_{k+1}.$$

By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.