Sum of an infinite series of fractions I am taking a fun walk into Number Theory land, and I am conducting an investigation about an infinite sum of fractions. These fractions have to do with the amount of composites within an even number. All the fractions have to do with the reciprocal of a prime.
This is my sum:
$\frac{1}{3} + \frac{1}{5} - (\frac{1}{3}*\frac{1}{5}) + \frac{1}{7}-(\frac{1}{3} * \frac{1}{7} + \frac{1}{5}*\frac{1}{7}) + \frac{1}{11} - (\frac{1}{3}*\frac{1}{11} + \frac{1}{5} * \frac{1}{11} + \frac{1}{7} * \frac{1}{11}) + ...  $
I don't quite know how to simplify the sequence, or even whether it is convergent or divergent.
Each term $ a_n $ in the series involves the next largest prime.
$ a_n = \frac{1}{p_n} - (\frac{1}{p_{n-1}}*\frac{1}{p_n} + \frac{1}{p_{n-2}}*\frac{1}{p_n} + \ \   ...)$
Therefore,
$\frac{1}{7}-(\frac{1}{3} * \frac{1}{7} + \frac{1}{5}*\frac{1}{7})$
is a single term, where 7 would be $p$ and 5 would be $p_{n-1}$ and 3 would be $p_{n-2}$.
EDIT:
I have been having trouble understanding some of the answers (I understand the math, just not the answer). The basis for the problem was the following: if 1/3 of all positive integers are divisible by 3, and 1/5 of all positive integers are divisible by 5, and there is an overlap here of 1/15, ($\frac{1}{3} * \frac{1}{5}$), then $\frac{1}{3} + \frac{1}{5} - \frac{1}{15}$ represents all numbers that are divisible by either 3 or 5. I expanded this to include all prime  numbers in my series, and was hoping that someone would show that the series would converge to 1, allowing me to continue on my way with a new piece of knowledge.
The proofs below show that this sum goes to negative infinity--that is, the amount of composite numbers expressed as some number between 0 and 1 goes to negative infinity. How is this possible? (please provide an explanation in your answer or update it) Maybe I have made some sort of logical or conceptual mistake that needs to be addressed, or maybe there is actually a good way of interpreting this answer. Thanks!
 A: The partial sums can be rearranged as
$$\sum_{k=2}^n a_k = \sum_{k=2}^n \frac{1}{p_k}\left(1 - \sum_{j=2}^{n-1} \frac{1}{p_j}\right)
= \sum_{k=2}^n \frac{1}{p_k} - \sum_{2\le j < k \le n}\frac{1}{p_jp_k}\\
= \sum_{k=2}^n \frac{1}{p_k} - \frac12\left[ \left(\sum_{k=2}^n \frac{1}{p_k}\right)^2 - \left(\sum_{k=2}^n \frac{1}{p_k^2}\right)\right] 
$$
It is known that following limits exists
$$\lim_{N\to\infty}\left(\sum_{p_k \le N} \frac{1}{p_k} - \log\log N\right)
= \gamma + \sum_{k=1}^\infty\left[\log\left(1 - \frac{1}{p_k}\right) + \frac{1}{p_k} \right]$$
(known as Mertens constant).
Together with Prime number theorem which essentially say
$p_n \sim n \log n$ for large $n$ and following bound:
$$\sum_{k=2}^n \frac{1}{p_k^2} \le \sum_{k=2}^n \frac{1}{k^2} < \frac{\pi^2}{6} - 1 < \infty$$
We obtain
$$\sum_{k=2}^n \frac{1}{p_k} \sim O(\log\log(n\log n))
\quad\implies\quad
\sum_{k=2}^n a_k \sim O((\log\log(n\log n))^2)$$
The series at hand diverges to $-\infty$ very slowly.
Update 
About the updated question why the answer can be negative.
Let's look at number $105 = 3\times 5 \times 7$ as an example,


*

*you counted it 3 times, in $\frac13$, $\frac15$ and $\frac17$.

*you cancelled it 3 times, in $\frac13\times\frac15$, $\frac13\times\frac17$ and $\frac15\times\frac17$.


This means you are not counting $105$ at all. 
In general, for any number that contain $m \ge 3$ distinct prime factors, you
over cancel it $\frac{m(m-1)}{2} - (m - 1) = \frac{(m-1)(m-2)}{2}$ times. 
Even though the fraction of numbers having a specific $m$ getting smaller and smaller as $m$ increases, it does not drop fast enough to suppress this $O(m^2)$ dependence. 
At the end, you over cancel too much and the result diverges to $-\infty$.
To properly deal with this sort of over counting, you need to use
Inclusion-exclusion principle.
The expression you should use should be something like:
$$\sum_{2\le i_1\le n} \frac{1}{p_{k_1}}
+ \sum_{2\le i_1<i_2 \le n} \frac{(-1)}{p_{i_1} p_{i_2}}
+ \sum_{2\le i_1<i_2<i_3 \le n} \frac{(-1)^2}{p_{i_1} p_{i_2} p_{i_3}}
+ \cdots
+ \sum_{2\le i_1<\cdots<i_{n-1}\le n} \frac{(-1)^{n-2}}{p_{i_1} \cdots p_{i_{n-1}}}
$$
This can be rewritten as
$\displaystyle\;1 - \prod_{k=2}^n\left(1 - \frac{1}{p_k}\right)\;$
and it does converge to $1$ as $n \to \infty$.
A: Using a little less firepower than achille hui's excellent answer, note that
$${1\over3}+{1\over5}+\cdots+{1\over23}+{1\over29}\approx1.0334\gt1+{1\over100}$$
and thus
$$1-{1\over3}-{1\over5}-\cdots-{1\over p}\lt{-1\over100}\quad\text{if }p\ge29$$
Using $S_{29}$ to denote the early part of the sum, the OP's series can be rewritten as
$$S_{29}+{1\over31}\left(1-{1\over3}-{1\over5}-\cdots-{1\over 29}\right)+{1\over37}\left(1-{1\over3}-{1\over5}-\cdots-{1\over 31}\right)+\cdots\\\lt S_{29}-{1\over100}\left({1\over31}+{1\over37}+{1\over41}+\cdots\right)$$
Since the sum of the reciprocals of the primes diverges to $\infty$, the OP's series diverges to $-\infty$.
Update (in response to OP's additional question):  I see what you're getting at.  Roughly speaking, you want to think of your series as computing the fraction of the integers that are not pure powers of $2$, the answer to which would be $1$.  That is, $1\over3$ of the integers are multiples of $3$; an additional $1\over5$ are multiples of $5$, but that double counts the multiples of $15$, so one should subtract a fraction ${1\over3}\cdot{1\over5}$.  And so on.
The fly falls into the ointment here at the next prime.  It makes perfect sense, initially, to subtract the fractions ${1\over3}\cdot{1\over7}$ and ${1\over5}\cdot{1\over7}$.  But then you need to realize that this subtracts too much:  In adjusting downward for the multiples of $21$ and $35$, you've adjusted downward twice for the multiples of $105$.  This over-subtraction persists, and gets worse, as you go further out in the series.  
Hopefully this helps explain why the series winds up diverging to $-\infty$.
