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?
 A: Such regrouping of a series is just not guaranteed not to alter the limit or even preserve convergence. 
It would be admissible if the series were absolutely convergent, that is the series of absolute values would converges. But, the current series does not converge absolutely. 
A: The series $\sum\limits_{k=1}^n\frac{(-1)^{k+1}}k$ is convergent but not absolutely convergent, i.e. the sum itself has a limit, but $\sum_{k=1}^\infty\left\lvert\frac{(-1)^{k+1}}{k}\right\rvert=+\infty$. Therefore, by Riemann-Dini theorem, for any extended real numbers $\alpha\le \beta\in\Bbb R\cup\{-\infty,\infty\}$ there is a bijective function $f:\Bbb N^+\to\Bbb N^+$ such that $$\alpha=\liminf_{n\to\infty}\sum_{k=1}^n \frac{(-1)^{f(k)}}{f(k)}\le\limsup_{n\to\infty}\sum_{k=1}^n \frac{(-1)^{f(k)}}{f(k)}=\beta$$
You happen to have described, more or less explicitly, a bijective function $f$ which goes well with $\alpha=\beta=\ln\sqrt2$, which happens to be the example(s) on Wikipedia too.
A: 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 $1$.
A: In your new series
$$\left(\frac 11 - \frac 12\right) - \frac 14 + 
\left(\frac 13 - \frac 16\right) - \frac 18 + 
\left(\frac 15 - \frac 1{10}\right) - \frac 1{12} +
\left(\frac 17 - \frac 1{14}\right) - \frac 1{16} \dots$$
Terms of the form $\frac{1}{2z - 1}$ occur 1/3 of the time instead of 1/2 the time like they do in the original series, so they are underweighted.
Terms of the form $\frac{1}{2z}$ occur 2/3 of the time instead of 1/2 the time like they do in the original series, so they are overweighted.
By changing the density at which those terms appear, you are effectively converting the sum from
$$\lim_{n \to \infty} \sum_{k = 1}^n \frac{1}{2k - 1} - \sum_{k = 1}^n \frac{1}{2k}$$
into 
$$\lim_{n \to \infty} \sum_{k = 1}^{n \cdot 2/3} \frac{1}{2k - 1} - \sum_{k = 1}^{n \cdot 4/3} \frac{1}{2k}$$
which naturally gives a smaller value since the subtracted terms have increased density.  If instead you rearrange the series without changing the density of the subseries, then you'll get an equal result.
A: You may not rearrange the series if it's not absolutely convergent. Otherwise you can get results like these, and like $\sum_{n=0}^\infty (-1)^k = \frac{1}{2}$, which both are false (the last one being the famous Grandi's series, https://en.wikipedia.org/wiki/Grandi%27s_series). All of this is in the Riemann series theorem (https://en.wikipedia.org/wiki/Riemann_series_theorem). Basically, that's the reason why your result is incorrect, and I hope this helps you!
