I tried to solve this problem using mathematical induction. First, we say that $ 3 \lt 4$ for $n =2$, which is correct. Then by assuming that it's accurate for $k$, we need to prove it's accurate as well for each $ k+1$. $$p_{k+1} \lt 2^{k+1} \Rightarrow p_{k+1} \lt 2^{k} \times 2 \Rightarrow \cfrac{p_{k+1}}{2} \lt 2^{k} $$ I don't know how should I prove it at last and I don't know if this is the way of solving this or not. Can someone correct me if I'm wrong and help me figure out the rest?
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4https://en.wikipedia.org/wiki/Bertrand's_postulate – Arthur Oct 31 '21 at 09:49
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1This is not exactly what I would call a problem for beginners. – José Carlos Santos Oct 31 '21 at 09:54
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1There is an even lower bound (that has been proven): https://math.stackexchange.com/questions/206815/is-there-a-way-to-show-that-sqrtp-n-n – Oct 31 '21 at 10:03
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@JoséCarlosSantos Unless say the context is a class where they've covered whatshisname's $n<p<2n$ thing, and the exercise is just to recognize the relevance... – David C. Ullrich Oct 31 '21 at 10:44
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It's even more surprising. Use it as an example ... – rtybase Oct 31 '21 at 10:45