Question Number 223224 by wewji12 last updated on 18/Jul/25
$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{k}}=\infty \\ $$$$\underset{\mathrm{p}\:\mathrm{prime}} {\sum}\:\frac{\mathrm{1}}{{p}}=\infty \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}….\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}+…\right)=?? \\ $$
Commented by Ghisom last updated on 18/Jul/25
![just a logical argument: there are more non−primes than primes ⇒ Q>P but P=∞ ⇒ Q=∞ [P=Σ_(p prime) (1/p) ∧ Q=Σ_(q not prime) (1/q)] Q_n =H_n −P_n H_(100) ≈5.19 P_(100) ≈2.11 Q_(100) ≈3.08](https://www.tinkutara.com/question/Q223237.png)
$$\mathrm{just}\:\mathrm{a}\:\mathrm{logical}\:\mathrm{argument}: \\ $$$$\mathrm{there}\:\mathrm{are}\:\mathrm{more}\:\mathrm{non}−\mathrm{primes}\:\mathrm{than}\:\mathrm{primes} \\ $$$$\Rightarrow\:{Q}>{P} \\ $$$$\mathrm{but}\:{P}=\infty\:\Rightarrow\:{Q}=\infty \\ $$$$\left[{P}=\underset{{p}\:\mathrm{prime}} {\sum}\frac{\mathrm{1}}{{p}}\:\wedge\:{Q}=\underset{{q}\:\mathrm{not}\:\mathrm{prime}} {\sum}\frac{\mathrm{1}}{{q}}\right] \\ $$$$ \\ $$$${Q}_{{n}} ={H}_{{n}} −{P}_{{n}} \\ $$$${H}_{\mathrm{100}} \approx\mathrm{5}.\mathrm{19} \\ $$$${P}_{\mathrm{100}} \approx\mathrm{2}.\mathrm{11} \\ $$$${Q}_{\mathrm{100}} \approx\mathrm{3}.\mathrm{08} \\ $$