What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:

Is  $\ln(2)=\frac{1}{2}\ln(2)$??

The following argument seems suggesting that the answer is yes:

We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
  which has a mathematically determined value $\ln(2)=0.693$. 
Now, let's do some rearrangement:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......$$
   $$
 (1-\frac{1}{2})-\frac{1}{4}+(\frac{1}{3}-\frac{1}{6})-\frac{1}{8}+(\frac{1}{5}-\frac{1}{10})-\frac{1}{12}.......$$
  $$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}......$$
   $$\frac{1}{2}(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}......)$$
$$\frac{1}{2}\ln(2).$$


I know that mathematics can't be wrong, and I have done something wrong. But here is my question: where does the argument above go wrong?

As others have pointed out, rearrangement is not allowed, so I will give you the most extreme case:
$$S=1-\frac12-\frac14-\frac16-\frac18+\frac13+\frac15+\frac17+\frac19+\frac1{11}+\frac1{13}+\dots$$
The general pattern is simple.  Start at $1$.  Then add up the even terms until the sum is less than $0$.  Then add up the odd terms until the sum is more than $1$.  Then add up the even terms until the sum is less than $0$ again...
Clearly if we repeat this process indefinitely, we have rearranged it so that the sum never converges and oscillates between $0$ and $1$.
