${\displaystyle \underbrace {\sum _{k=0}^{N-1}{\frac {2k+1}{z^{2k+1}-1}}} _{A_{N}}+\underbrace {\sum _{k=1}^{N}{\frac {2k}{z^{2k}-1}}} _{B_{N}}=\underbrace {\sum _{k=1}^{2N}{\frac {k}{z^{k}-1}}} _{C_{2N}}}$.

Wegen ${\displaystyle {\frac {2k}{z^{2k}-1}}={\frac {k}{z^{k}-1}}-{\frac {k}{z^{k}+1}}}$ ist

${\displaystyle B_{N}=\underbrace {\sum _{k=1}^{N}{\frac {k}{z^{k}-1}}} _{C_{N}}-\underbrace {\sum _{k=1}^{N}{\frac {k}{z^{k}+1}}} _{D_{N}}}$.

Also ${\displaystyle A_{N}=D_{N}+C_{2N}-C_{N}\,}$.

Für ${\displaystyle N\to \infty \,}$ folgt daraus die Behauptung.

Definiert man ${\displaystyle a_{n}\,}$ als ${\displaystyle {\frac {2^{n}}{n}}\sum _{k=0}^{n-1}{\frac {1}{n-1 \choose k}}}$,

so ist ${\displaystyle a_{n+1}-a_{n}={\frac {2^{n+1}}{n+1}}+\sum _{k=0}^{n-1}\underbrace {\left[{\frac {2^{n+1}}{n+1}}\,{\frac {1}{n \choose k}}-{\frac {2^{n}}{n}}\,{\frac {1}{n-1 \choose k}}\right]} _{=:b_{k}}}$.

Wegen ${\displaystyle {n \choose k}={\frac {n}{n-k}}\,{n-1 \choose k}}$ ist ${\displaystyle b_{k}={\frac {2^{n}}{n(n+1)}}\,{\frac {n-1-2k}{n-1 \choose k}}}$,

und daher ${\displaystyle b_{n-1-k}=-b_{k}\,}$. Also muss ${\displaystyle \sum _{k=0}^{n-1}b_{k}=0}$ sein.

Und somit ist ${\displaystyle a_{n+1}-\underbrace {a_{0}} _{=0}=\sum _{k=0}^{n}(a_{k+1}-a_{k})=\sum _{k=1}^{n+1}{\frac {2^{k}}{k}}}$.