# A series of reciprocal of integers [closed]

Let $F_n$ be integers, and $F_1<F_2<\cdots<F_n<\cdots$. Suppose that $$\lim_{n\to\infty}\frac{F_1F_2\cdots F_{n-1}}{F_n}=0.$$ Prove then $$\sum_{n=1}^\infty \frac{1}{F_n}$$ is convergence, and the sum of which is irrational.

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## closed as off-topic by user7530, Lord Soth, Old John, M Turgeon, Brandon CarterOct 30 '13 at 3:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user7530, Lord Soth, Old John, M Turgeon, Brandon Carter
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This is a PSQ, but I think it is an interesting one. Why close this, as opposed to some of the other incredible PSQs we've had in the past? (I, for one, think this should remain open) –  anorton Oct 30 '13 at 2:47
@anorton No self-work and "Really difficult, isn't it?" might have been annoying to some. –  Lord Soth Oct 30 '13 at 2:57
@LordSoth I agree that the "Really difficult..." comment made me want to close the question at first, but then I realized that Stack Exchange is really more about the question than the particular phrasing of the asker. (Because I could go and edit the "Really difficult" comment away right now... :) ) –  anorton Oct 30 '13 at 2:58
Why is this question on hold? Can't it be done by modifying the proof to show e is irrational but ending up with an integer equaling the product of two terms both going to 0 and thus is <1. –  domoremath Oct 31 '13 at 3:26

## 1 Answer

There are many exceptionally slow sequences, consider ${\prod_{i=1}^{n-1}F_i\over F_n}=\frac 1{\ln n}$ or even slower versions. But whatever sequence $1\over f(n)$ is chosen for the RHS, assuming $F_1$ is positive, we have $f(n)\to \infty$ as $n\to \infty$, thus

$$F_n=f(n)\prod_{i=1}^{n-1}F_i\ge \prod_{i=1}^{n-1}F_i$$

Then, assuming $\exists i:F_i\ge 2$, for all but a finite set of terms, we have

$$\sum_{i=k}^\infty\frac 1{F_i}\le \sum_{i=1}^\infty\frac 1{2^{2i}}$$

And thus $\sum_{i=1}^\infty \frac 1{F_i}$ converges under the specified assumptions.

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