[itex] <br /> <a href="https://www.physicsforums.com/insights/explore-the-fascinating-sums-of-odd-powers-of-1-n/" class="link link--internal">Continue reading...</a>[/itex]The goal is to get a little bit closer to the values of the zeta function (ζ(s)) and the eta function (η(s)) for some odd values of s. This insight is an expansion of two of my previous insights (Further Sums Found Through Fourier Series, Using the Fourier Series To Find Some Interesting Sums). You patience is appreciated as there are many equations to load on this page.
Specifically I will calculate the sums
[itex]\frac{1}{1^{p}}-\frac{1}{3^{p}} +\frac{1}{5^{p}}-\frac{1}{7^{p}}…[/itex]
[itex]\frac{1}{1^{p}}+\frac{1}{2^{p}}-\frac{1}{4^{p}} -\frac{1}{5^{p}}+\frac{1}{7^{p}}…[/itex]
[itex]\frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}} -\frac{1}{5^{p}}-\frac{1}{6^{p}}-\frac{1}{7^{p}}…[/itex]
[itex]\frac{1}{1^{p}}+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}} +\frac{1}{5^{p}}-\frac{1}{7^{p}}-\frac{1}{8^{p}}…...[/itex]
Geoffrey Campbell
I think that Euler had already found these results by utilizing the infinite product for (sin x)/x and it's variations starting from an heuristic argument about the roots of an infinite polynomial (the power series in fact). The approach taken in this article is more modern and uses the fact of contemporary calculations of the coefficients, which is nowadays easy. A nice slant on this, however, unfortunately not giving us the desired zeta values for the odd numbers 3, 5, 7, etc.
No, it is not a misprint. The eta function resembles the zeta function, the difference lies in the sign of the even powers of 1/n (https://en.wikipedia.org/wiki/Dirichlet_eta_function).A. Neumaier said:In the line before Section 5 you mention the eta function, presumably a misprint...