Ivan Shishkin, Rye (1878)

Problems/Real analysisUnreviewed

Divergence of ∑1p\sum \frac{1}{p}

by mathman·translated by visitor·
30
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This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

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Show that the sum
∑p⩽n1p,\sum_{p \leqslant n} \frac{1}{p},where pp runs over the primes, diverges.

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Solution by mathmanFR

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∑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))

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