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

let p_{i} be the i^{th} prime

Does the fillowing sum converge :

\underset{i = 1}{\sum^{\infty}} \frac{p_{i}}{p_{i + 1}}

(p_{1} = 2, p_{2} = 3, p_{3} = 5, \dots)

Answered by 123456 last updated on 19/Dec/15

let

A = \underset{i = 1}{\sum^{\infty}} \frac{p_{i}}{p_{i + 1}}

we have

\frac{p_{i}}{p_{i + 1}} ⩾ \frac{1}{p_{i + 1}}

so

A = \underset{i = 1}{\sum^{\infty}} \frac{p_{i}}{p_{i + 1}} ⩾ \underset{i = 1}{\sum^{\infty}} \frac{1}{p_{i + 1}} = \sum_{p \in P} \frac{1}{p} - \frac{1}{2}

since the sun of inverse of primes diverge

then A diverge

Commented by Filup last updated on 19/Dec/15

Nice work!!!!

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