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Question Number 223049 by wewji12 last updated on 13/Jul/25
we already know Σ_(h=1) ^∞ (1/h)=∞ and Σ_(p∈P) (1/p)=∞ from the Harmonic Series Σ_(h=1) ^∞ (1/h)=Π_(p∈P) (1+(1/p)+((1/p))^2 +((1/p))^3 +...)= Π_(p∈P) (1/(1−p^(−1) )) because ζ(s)=Π_(p∈P) (1/(1−p^(−s) )) ζ(1)=Σ_(h=1) ^∞ ((1/h))^1 =Π_(p∈P) (1/(1−p^(−1) )) ln(Σ_(h=1) ^∞ (1/h))=ln(Π_(p∈P) ^ (1/(1−p^(−1) )))=−Σ_(p∈P) ln(1−(1/p)) Σ_(p∈P) ((1/p)+(1/2)((1/p))^2 +(1/3)((1/p))^3 +...)= Σ_(p∈P) (1/p)+(1/2)Σ_(p∈P) ((1/p))^2 +(1/3)Σ_(p∈P) ((1/p))^3 +........ by using the following relation Σ_(h=1) ^∞ (1/h)=∞ ln((1/(1−z)))=Σ_(h=1) ^∞ (z^h /h) , z→(1/p_j ) Σ_(j=1) ^∞ ln(1−(1/p_j ))=Σ_(j=1) ^∞ Σ_(h=1) ^∞ (1/h)((1/p_j ))^h Σ_p ln(1−(1/p))=ln(Σ_(h=1) ^∞ (1/h))=Σ_(p∈P) Σ_(h=1) ^∞ (1/h)((1/p))^h ∴Σ (1/p) is div. my Question is Σ_(k=1) ^∞ (((−1)^k )/p_k ) =conv???
$$\mathrm{we}\:\mathrm{already}\:\mathrm{know}\:\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\infty\:\mathrm{and}\:\underset{{p}\in\mathbb{P}} {\sum}\:\frac{\mathrm{1}}{{p}}=\infty \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{Harmonic}\:\mathrm{Series} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\underset{{p}\in\mathbb{P}} {\prod}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}+\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{3}} +…\right)=\:\underset{{p}\in\mathbb{P}} {\prod}\:\frac{\mathrm{1}}{\mathrm{1}−{p}^{−\mathrm{1}} } \\ $$$$\mathrm{because}\:\zeta\left({s}\right)=\underset{{p}\in\mathbb{P}} {\prod}\:\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} } \\ $$$$\zeta\left(\mathrm{1}\right)=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\frac{\mathrm{1}}{{h}}\right)^{\mathrm{1}} =\underset{{p}\in\mathbb{P}} {\prod}\:\frac{\mathrm{1}}{\mathrm{1}−{p}^{−\mathrm{1}} } \\ $$$$\mathrm{ln}\left(\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}\right)=\mathrm{ln}\left(\underset{{p}\in\mathbb{P}} {\overset{\:} {\prod}}\:\frac{\mathrm{1}}{\mathrm{1}−{p}^{−\mathrm{1}} }\right)=−\underset{{p}\in\mathbb{P}} {\sum}\:\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}}\right) \\ $$$$\underset{{p}\in\mathbb{P}} {\sum}\:\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{3}} +…\right)= \\ $$$$\underset{{p}\in\mathbb{P}} {\sum}\:\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{p}\in\mathbb{P}} {\sum}\:\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\underset{{p}\in\mathbb{P}} {\sum}\:\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{3}} +…….. \\ $$$$\mathrm{by}\:\mathrm{using}\:\mathrm{the}\:\mathrm{following}\:\mathrm{relation} \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}=\infty \\ $$$$\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}−{z}}\right)=\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{z}^{{h}} }{{h}}\:\:,\:{z}\rightarrow\frac{\mathrm{1}}{{p}_{{j}} } \\ $$$$\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\:\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}_{{j}} }\right)=\underset{{j}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}\left(\frac{\mathrm{1}}{{p}_{{j}} }\right)^{{h}} \\ $$$$\underset{{p}} {\sum}\:\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}}\right)=\mathrm{ln}\left(\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}\right)=\underset{{p}\in\mathbb{P}} {\sum}\:\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{h}}\left(\frac{\mathrm{1}}{{p}}\right)^{{h}} \\ $$$$\therefore\Sigma\:\frac{\mathrm{1}}{{p}}\:\mathrm{is}\:\mathrm{div}. \\ $$$$\mathrm{my}\:\mathrm{Question}\:\mathrm{is}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{p}_{{k}} }\:=\mathrm{conv}??? \\ $$

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