Let $p_{k}$ be the $k^{th}$ prime. Show that $p_{k}≥2k-1$ for all $k≥2$. This inequality is true for several values of $k$. For example $p_{2}=3≥2(2)-1=3$ and $p_{3}=5≥2(3)-1=5$.
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PNT or are you looking for a simpler solution? – rtybase Oct 05 '19 at 16:53
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@rtybase: A simpler solution. – Safwane Oct 05 '19 at 16:54
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3Induction? $p_k\geq 2k-1 \Rightarrow p_{k+1}\geq p_k+2\geq 2(k+1)-1$ – rtybase Oct 05 '19 at 16:55
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Hint: No even positive integers are prime except for $2$. Alternatively, as rtybase's comment suggested, you can also use mathematical induction.
John Omielan
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