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If we iteratively remove the multiples of the succesive prime numbers from the natural numbers, starting from 2, i.e.,

$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... \rightarrow$

$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... \rightarrow$

$1, 5, 7, 11, 13, 17, 19, 23, 25, ... \rightarrow$

we are consecutively removing fractions

$$\frac{1}{2}, \; \frac{1}{2·3}, \; \frac{1}{2·3·5}, \; \frac{1}{2·3·5·7},\; ... $$

from the total amount of natural numbers. This suggests that the infinite sum

$$\frac{1}{2}+ \frac{1}{2·3}+ \frac{1}{2·3·5}+ \frac{1}{2·3·5·7}+ ...$$

is equal to 1, but I couldn't figure out a proof for this result. Some references about this particular sum would be appreciated too. Thank you in advance

JorgePV
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2 Answers2

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This is not the case: $$\frac12+\frac1{2\cdot 3}+\frac1{2\cdot 3\cdot 5}+\frac1{2\cdot 3\cdot 5\cdot 7}+\frac1{2\cdot 3\cdot 5\cdot 7\cdot11}+\cdots \le \frac12\left(1+\frac1{3}+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\cdots \right)=\frac34$$

It seems that your series converges to $$0.70523017179180096514743168288824851374357763910915432819226791381\ldots$$ and does so way faster than the geometric series used in the above comparison. After all, the $n$th summand is $$\le \frac12\cdot\frac{2^n\cdot n!}{(2n)!}\approx \frac{2^nn^ne^{-n}\sqrt{2\pi n}}{2\cdot(2n)^{2n}e^{-2n}\sqrt{4\pi n} }=\frac1{2\sqrt 2}\cdot \left(\frac e{2n}\right)^n.$$

Raffaele
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Hagen von Eitzen
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  • I see, thank you! I just can't see now where my approach failed, as it seems to make sense. I will think further about it – JorgePV Nov 27 '20 at 14:57
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The denominators are called primorials, the series have been studied and proved irrational in this article.

jjagmath
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