A classic example of the Riemann series theorem use the alternating harmonic series $\sum_{n=1}^k \frac{(-1)^{n+1}}{n}$ that converges to $\ln(2)$ when $k \rightarrow \infty $ and rearrange it to $\sum_{n=1}^k (\frac{1}{2n - 1} - \frac{1}{4n - 2} - \frac{1}{4n})$ that converges to $\frac{1}{2}\ln(2)$ when $k \rightarrow \infty $
But if we strictly compare the terms for those two sum from n=1 to k with $\displaystyle \sum_{n=1}^{4k} \frac{(-1)^{n+1}}{n}$ we found for k=1 :
\begin{array}{l|cccc} \displaystyle \sum_{n=1}^4 \frac{(-1)^{n+1}}{n} & 1 & -\frac{1}{2} & +\frac{1}{3} & -\frac{1}{4} \\ \hline \displaystyle \sum_{n=1}^1 (\frac{1}{2n - 1} - \frac{1}{4n - 2} - \frac{1}{4n}) & 1 & -\frac{1}{2} & & -\frac{1}{4} \\ \hline \small \text{Term not taken in account} & & & +\frac{1}{3} \\ \small \text{in the rearranged series} \end{array}
for k=2 :
\begin{array}{l|cccc} \displaystyle \sum_{n=1}^8 \frac{(-1)^{n+1}}{n} & 1 & -\frac{1}{2} & +\frac{1}{3} & -\frac{1}{4} & +\frac{1}{5}& -\frac{1}{6} & +\frac{1}{7}& -\frac{1}{8} \\ \hline \displaystyle \sum_{n=1}^2 (\frac{1}{2n - 1} - \frac{1}{4n - 2} - \frac{1}{4n}) & 1 & -\frac{1}{2} & +\frac{1}{3} & -\frac{1}{4} & & -\frac{1}{6} & & -\frac{1}{8} \\ \hline \small \text{Terms not taken in account} & & & & & +\frac{1}{5} & & +\frac{1}{7} \\ \small \text{in the rearranged series} \end{array}
for k=3 :
\begin{array}{l|cccc} \displaystyle \sum_{n=1}^{12} \frac{(-1)^{n+1}}{n} & 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} \\ \hline \sum_{1}^3 (\frac{1}{2n - 1} - \frac{1}{4n - 2} - \frac{1}{4n}) & 1 -\frac{1}{2} +\frac{1}{3} -\frac{1}{4} +\frac{1}{5} -\frac{1}{6} & & -\frac{1}{8} & & -\frac{1}{10}& & -\frac{1}{12} \\ \hline \small \text{Terms not taken in account} & & +\frac{1}{7} & & +\frac{1}{9} & & +\frac{1}{11} \\ \small \text{in the rearranged series} \end{array}
And so on, with always a number of terms not taken in account in the rearranged series increasing with k, which are the inverses of odds numbers going from $\frac{1}{2k + 1}$ to $\frac{1}{4k-1}$ which is equal to $\sum_{p=k}^{2k-1}\frac{1}{2p+1}$ (thanks to Carlo) which can be approximated by an integral over p equals to $\frac{1}{2}\ln(\frac{4k−1}{2k+1})$ that converges to $\frac{1}{2}\ln(2)$ when $k \rightarrow \infty $
In short :
\begin{array}{lllll} \displaystyle \sum_{n=1}^k (\frac{1}{2n - 1} - \frac{1}{4n - 2} - \frac{1}{4n}) & = & \displaystyle \sum_{n=1}^{4k} \frac{(-1)^{n+1}}{n} & - & \displaystyle \sum_{p=k}^{2k-1}\frac{1}{2p+1} \\ \small \text{Rearranged series} & = & \small \text{original series} & - & \small \text{Terms not taken in account} \\ \frac{1}{2}\ln(2) & = & \ln(2) & - & \frac{1}{2}\ln(2) \end{array}
So the Riemann series theorem is false ?