I really didn't know how to word the title so sorry if it's a little confusing.(adsbygoogle = window.adsbygoogle || []).push({});

And I didn't know whether to post this in number theory or not but ah well.

The other day, I started thinking about this and I was just wondering if it had been done before or if it's even correct;

Half of all positive integers are divisible by 2,

A third are divisible by 3,

A quarter by 4 etc.

Now, of the third that are divisible by 3, a half of those numbers are also divisible by 2.

So 1/6 of positive integers are divisible by 3 and not 2.

I continued this trend to find;

[tex]\frac{1}{2}[/tex] of p. ints are divisible by 2

[tex]\frac{1}{3}\left(1-\frac{1}{2}\right)[/tex] are divisible by 3 and not 2

[tex]\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right)[/tex] are divisible by 5 and not 3 or 2

[tex]\frac{1}{7}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)+\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right)\right)\right)[/tex] are divisible by 7 and not 5, 3 or 2

If I sum all these to infinity, it should equal 1 as it will then cover all positive integers.

So could you by any chance verify the truth of this statement,

[tex]\sum^{\infty}_{p_{0}=2}\frac{1}{p_{0}}\left(1-\sum^{p_{0}-1}_{p_{1}=2}\frac{1}{p_{1}}\left(1-\sum^{p_{1}-1}_{p_{2}=2}\frac{1}{p_{2}}\left(...\right)\right)\right) = 1[/tex]

I know this is quite awkward looking and I don't even know if I've used correct notation, I just sort of scrawled it down as I was thinking it.

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# Fractions of all numbers

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