Over \(\sum\limits_{n=1}^m \dfrac{1}{n^{2h}}\) en \(\sum\limits_{n=1}^\infty \dfrac{1}{n^{2h}}= \zeta (2h)\).

zbMath55.0130.06MaRDI QIDQ1835050

L. W. Nieland

Publication date: 1929

Published in: Nieuw Archief voor Wiskunde. Tweede Serie (Search for Journal in Brave)

This page was built for publication: Over \(\sum\limits_{n=1}^m \dfrac{1}{n^{2h}}\) en \(\sum\limits_{n=1}^\infty \dfrac{1}{n^{2h}}= \zeta (2h)\).