<s>[INST] 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?
 [/INST] 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.
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