Is Every Positive Integer Divisible by 2, 3, or 5?

[marcusmath]

I really didn't know how to word the title so sorry if it's a little confusing.
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;
\frac{1}{2} of p. ints are divisible by 2
\frac{1}{3}\left(1-\frac{1}{2}\right) are divisible by 3 and not 2
\frac{1}{5}\left(1-\left(\frac{1}{2}+\frac{1}{3}\left(1-\frac{1}{2}\right)\right)\right) are divisible by 5 and not 3 or 2
\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) 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,

\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

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.

 

[CRGreathouse]

To make it a little easier: the sum of "divisible by 2" + "not divisible by 2 but divisible by 3" + "not divisible by 2 or 3 but divisible by 5" is "divisible by 2, 3, or 5" which is "not relatively prime to 2, 3, and 5":

f_1 = 1/2
f_2 = 1/2 + 1/2\cdot1/3 = 1 - 1/2\cdot2/3
f_3 = 1/2 + 1/2\cdot1/3 + 1/2\cdot2/3\cdot1/5 = 1 - 1/2\cdot2/3\cdot4/5
. . .
f_k = 1 - \sum_{q\in\mathbb{P},q\le p_k}\left(1-\frac1p\right)

Now instead of lots of sums, you only have one. All you need to show is that the sum in question tends to 0 as k goes to infinity (which is true).