Let's consider the alternating harmonic series
$S_n = 1-\frac12 + \frac13 – \cdots + (-1)^n\frac1n$.
By rearranging its terms, we get
$S_n = (1-\frac12)-\frac14 + (\frac13-\frac16)-\frac18 + (\frac15-\frac1{10})-\cdots$.
This equals to $S_n = \frac12-\frac14 + \frac16-\frac18 + \frac1{10}-\cdots$.
By extracting $\frac12$ as common factor, we get:
$S_n = \frac12(1-\frac12 + \frac13-\frac14 + \cdots)$.
So in essence, $S_n = \frac12 S_n$,
therefore $1=\frac12$.
I have read the wikipedia article about Riemann series and roughly my understanding is that if the series converges, we can rearrange the terms and get any other number, or even to diverge.
What could be an acceptable explanation of the paradox? Obviously 1 does not equal $\frac12$! In which of the above steps lies the error?
Best Answer
It seems you are only considering partial sums $S_n$ hence the considerations of (absolute) convergence and rearrangements are offtopic.
However, if you make the "⋯" in your post explicit, you will see that one does not reach "in essence" $S_n=\frac12S_n$. For example, $$S_{10}=1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19-\frac1{10}$$ hence the reordering of terms that you suggest yields $$S_{10}=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\color{red}{0}+\left(\frac17-\color{red}{0}\right)-\color{red}{0}+\left(\frac19-\color{red}{0}\right)-\color{red}{0}$$ whose value is not at all equal to $$\frac12S_{10}=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\color{red}{\frac1{12}}+\qquad$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\left(\frac17-\color{red}{\frac1{14}}\right)-\color{red}{\frac1{16}}+\left(\frac19-\color{red}{\frac1{18}}\right)-\color{red}{\frac1{20}}$$ More generally, for every $n$, the reordering of the terms of $S_{2n}$ forgets $n$ terms in $\frac12S_{2n}$, which are equal to $-\color{red}{\frac1{2n+2k}}$ for $1\leqslant k\leqslant n$.
Edit: Likewise, $$S_9=1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\frac19$$ hence the suggested reordering yields $$S_9=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\color{red}{0}\right)-\color{red}{0}+\left(\frac17-\color{red}{0}\right)-\color{red}{0}+\left(\frac19-\color{red}{0}\right)$$ whose value is not at all equal to $$\frac12S_9=\left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\color{red}{\frac1{12}}+\qquad$$ $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\left(\frac17-\color{red}{\frac1{14}}\right)-\color{red}{\frac1{16}}+\left(\frac19-\color{red}{\frac1{18}}\right)$$ More generally, for every $n$, the reordering of the terms of $S_{2n+1}$ forgets $n$ terms in $\frac12S_{2n+1}$, which are equal to $-\color{red}{\frac1{2n+2k+2}}$ for $1\leqslant k\leqslant n$.