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Question Number 3685 by Filup last updated on 19/Dec/15

$$\mathrm{let}{p}_i\mathrm{be}\mathrm{the}{i}^{\mathrm{th}}\mathrm{prime}$$$$$$$$\mathrm{Does}\mathrm{the}\mathrm{fillowing}\mathrm{sum}\mathrm{converge}:$$$$\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{p_i}{p_{i+1}}$$$$(p_1=2,p_2=3,p_3=5,...)$$

Answered by 123456 last updated on 19/Dec/15

$$\mathrm{let}$$$$\mathrm{A}=\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{p_i}{p_{i+1}}$$$$\mathrm{we}\mathrm{have}$$$$\frac{p_i}{p_{i+1}}⩾\frac{1}{p_{i+1}}$$$$\mathrm{so}$$$$\mathrm{A}=\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{p_i}{p_{i+1}}⩾\underset{i=1}{\sum ^{\mathrm{\infty }}}\frac{1}{p_{i+1}}=\sum _{p\in \mathbb{P}}\frac{1}{p}-\frac{1}{2}$$$$\mathrm{since}\mathrm{the}\mathrm{sun}\mathrm{of}\mathrm{inverse}\mathrm{of}\mathrm{primes}\mathrm{diverge}$$$$\mathrm{then}\mathrm{A}\mathrm{diverge}$$

Commented by Filup last updated on 19/Dec/15

$$\mathrm{Nice}\mathrm{work}!!!!$$

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