本帖最后由 王守恩 于 2019-4-5 09:10 编辑
wayne 发表于 2018-7-3 20:39
指数为负的时候，级数是发散的。通过解析延拓可扩大定义范围（https://en.wikipedia.org/wiki/Analytic ...
谢谢wayne！ 左边不好算，右边好算。
\(公式(1)：\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{A}{A-1}\cdotn^{\frac{A-1}{A}}+\zeta\bigg(\frac{1}{A}\bigg)\ 譬如：\)
\(\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{2}{1}\cdotn^{\frac{1}{2}}+\zeta\bigg(\frac{1}{2}\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{3}{2}\cdotn^{\frac{2}{3}}+\zeta\bigg(\frac{1}{3}\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{4}{3}\cdotn^{\frac{3}{4}}+\zeta\bigg(\frac{1}{4}\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{5}{4}\cdotn^{\frac{4}{5}}+\zeta\bigg(\frac{1}{5}\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{6}{5}\cdotn^{\frac{5}{6}}+\zeta\bigg(\frac{1}{6}\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\frac{7}{6}\cdotn^{\frac{6}{7}}+\zeta\bigg(\frac{1}{7}\bigg)\)
\(\D\cdots\cdots\)
\(公式(2)：\D\sum_{k=1}^{n}\frac{1}{k^{A}}=\frac{A^{-1}}{A^{-1}-1}\cdotn^{\frac{A^{-1}-1}{A^{-1}}}+\zeta\bigg(A\bigg)\ 譬如：\)
\(\D\sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{2^{-1}}{2^{-1}-1}\cdotn^{\frac{2^{-1}-1}{2^{-1}}}+\zeta\bigg(2\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{k^{3}}=\frac{3^{-1}}{3^{-1}-1}\cdotn^{\frac{3^{-1}-1}{3^{-1}}}+\zeta\bigg(3\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{k^{4}}=\frac{4^{-1}}{4^{-1}-1}\cdotn^{\frac{4^{-1}-1}{4^{-1}}}+\zeta\bigg(4\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{k^{5}}=\frac{5^{-1}}{5^{-1}-1}\cdotn^{\frac{5^{-1}-1}{5^{-1}}}+\zeta\bigg(5\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{k^{6}}=\frac{6^{-1}}{6^{-1}-1}\cdotn^{\frac{6^{-1}-1}{6^{-1}}}+\zeta\bigg(6\bigg)\)
\(\D\sum_{k=1}^{n}\frac{1}{k^{7}}=\frac{7^{-1}}{7^{-1}-1}\cdotn^{\frac{7^{-1}-1}{7^{-1}}}+\zeta\bigg(7\bigg)\)
\(\D\cdots\cdots\)
王守恩 发表于 2019-4-5 09:06
谢谢wayne！ 左边不好算，右边好算。
我们可以这样说吗？
\(\D\sum_{k=2}^{n}\ \zeta k\ =n\)