SolutionSolution by mathman · FR∑p≤n1p>∑p≤nln(1+1p)=∑p≤n(ln(1−1p2)−ln(1−1p))>−ln(ζ(2))+ln(∑k=1n1k)>−ln(ζ(2))+ln(ln(n))\sum_{p\leq n}\frac{1}{p}> \sum_{p\leq n}ln(1+\frac{1}{p})=\sum_{p\leq n}(ln(1-\frac{1}{p²})-ln(1-\frac{1}{p}))> -ln(\zeta(2))+ ln(\sum_{k=1}^{n}\frac{1}{k})>-ln(\zeta(2))+ ln(ln(n))∑p≤np1>∑p≤nln(1+p1)=∑p≤n(ln(1−p21)−ln(1−p1))>−ln(ζ(2))+ln(∑k=1nk1)>−ln(ζ(2))+ln(ln(n))
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