An Infinite Series Contradiction Let A represent $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+......$ and B represent $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+......$
$$A+B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+......=2A-1$$
and then
$$A-B=1$$
thus 
$$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+......=1$$
Now set N as $\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+......$, a positive number
$$N+(A-B)=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+......=\frac{1}{2}$$
$$N+1=\frac{1}{2}$$
$$N=-\frac{1}{2}$$
What is wrong with this logic?
 A: I will try to point some things out, but I would strongly advise the OP to read about absolute convergence, conditional convergence and finally, summation of divergent series which they seem to be interested in.
First things first, as the comments pointed out, we can't manipulate with divergent series. They do not represent any real number and thus can not be added or subtracted to give a real number. We'd have the same luck adding the color blue and an apple and getting $\pi$ as a result.
A probably more striking example of such sensless games is this:
$$S=1+2+4+\dots+2^n+\dots \\ 2S=S-1 \\ S = - 1$$
This doesn't make sense, because adding all these terms we will never get the number on the right hand side (though again, I urge the OP to look up summation of divergent series).

Much more interesting topic is conditionally convergent series.
The most well known is the alternating harmonic series:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ \dots = \log 2$$
This one is not just formal, it really works. If we add the terms exactly in the same order as they are written. If we change the order of a finite number of terms, it's ok. Rules of arithmetic will hold. But if we change the order of the infinte number of terms, the value will change as well.
This can be seen by considering the general terms:
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$$
Since we are going in order, we can group the successive even and odd terms in pair, and that will not change the sum:
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\sum_{k=1}^\infty \left(\frac{1}{2k-1}-\frac{1}{2k} \right)=\sum_{k=1}^\infty \frac{1}{2k(2k-1)} $$
The general term is now always positive and the series we obtained converges absolutely, as can be seen from the denominator. Indeed, we can write:
$$\left(1-\frac{1}{2} \right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+ \dots = \log 2$$
In the same way we can group every $4$ terms, every $6$ terms and such, the value of the series will stay the same.

So what happens if we group the terms out of order? We will either get another value, or even get the series to diverge. Which is why they are called 'conditionally convergent'.
Let's now group the terms in another way:
$$\left(1-\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{6}-\frac{1}{8}\right)+\left(\frac{1}{5}-\frac{1}{10}-\frac{1}{12}\right)+ \dots = ?$$
To find the value of this new series (made of exactly the same terms with the same signs) we need to find the formula for the general term:
$$\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k}=\frac{1}{4k(2k-1)}$$
But we can see from the previous formula that:
$$\sum_{k=1}^\infty \frac{1}{4k(2k-1)}= \frac{\log 2}{2}$$
Somehow, we have obtained half the value of the original series.

Let's now try another way:
$$\left(1+\frac{1}{3}-\frac{1}{2}\right)+\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{4}\right)+\left(\frac{1}{9}+\frac{1}{11}-\frac{1}{6}\right)+ \dots = ?$$
$$\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}=\frac{8k-3}{2k(4k-3)(4k-1)}$$
Looking at the general term, we can see that the new series converges absolutely. Its value now:
$$\sum_{k=1}^\infty \frac{8k-3}{2k(4k-3)(4k-1)}= \frac{3\log 2}{2}$$


Unlike the OP's manipulations, all three series I obtained are absolutely convergent and really give us the values on the right hand side (or approximations to them, for a finite number of terms).

