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I apologize for being not able to provide much context.

Is there a special zeta function defined as: $$f(s)=\sum^\infty_{n=1}\frac1{(2n-1)^s}$$ ?

Moreover, if I know the value of $f(k)$($k$ is an integer), can we thus find the value of $\zeta(k)$?

I know very little about this topic, and recently am facing difficulties, so any help is highly appreciated.

Thanks in advance.

Szeto
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    In the "even" zeta function, we can factor $\frac{1}{2^s}$ out, writing it as $\frac{1}{2^s}\zeta(s)$, which means the "odd" zeta function is $\zeta(s)-\frac{1}{2^s}\zeta(s)=(1-\frac{1}{2^s})\zeta(s)$. (You can do something similar to relate the Riemann zeta function with the Dirichlet eta function, thus providing an analytic continuation of $\zeta(s)$ to all $s$ with ${\rm Re}(s)>0$ except $s=1$.) – anon May 17 '18 at 23:34
  • @anon Thank you. Maybe you can provide a detailed derivation of it as an answer. – Szeto May 17 '18 at 23:40

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\rule{0pt}{5mm}\mrm{f}\pars{s}\right\vert_{\ \Re\pars{s}\ >\ 1} & \equiv \sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{s}} = \sum_{n = 1}^{\infty}\bracks{{1 \over n^{s}} - {1 \over \pars{2n}^{s}}} = \pars{1 - 2^{-s}}\sum_{n = 1}^{\infty}{1 \over n^{s}} \\[5mm] & = \bbx{\pars{1 - 2^{-s}}\zeta\pars{s}} \end{align}

Felix Marin
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