Evaluating $\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{(2k+1)^{n+1}}$

Question

To prove an integral result, I have got to evaluate $$\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{(2k+1)^{n+1}} \text{ where } n\in\Bbb N$$ But since I haven't got much knowledge on evaluating series of this kind, some help or clues would be appreciated. Thank you in advance

Answer 1

One way to find the special values of the Dirichlet beta function at the odd natural numbers is to relate it to the Euler numbers through the Maclaurin series for the secant,

$$\sec(z)=\sum_{n=0}^\infty\frac{(-1)^n E_{2n}}{(2n)!}z^{2n},\qquad|z|<\pi/2.$$

If we take the logarithmic derivative of

$$\tan\left(\frac{\pi z}{2}\right)+\sec\left(\frac{\pi z}{2}\right)=\prod_{n=-\infty}^\infty\left(1-\frac{z}{4n-1}\right)\left(1-\frac{z}{4n+1}\right)^{-1}$$

we get

$$\begin{align*} \frac{\pi}{2}\sec\left(\frac{\pi z}{2} \right ) &=\sum_{n=-\infty}^\infty\left(\frac{-\frac{1}{4n-1}}{1-\frac{z}{4n-1}}-\frac{-\frac{1}{4n+1}}{1-\frac{z}{4n+1}} \right ) \\ &=\sum_{n=-\infty}^\infty\left(\sum_{k=0}^\infty\frac{z^k}{(4n+1)^{k+1}}-\sum_{k=0}^\infty\frac{z^k}{(4n-1)^{k+1}} \right ),\;\;|z|<\pi/2 \\ &\qquad\text{geometric series} \\ &=\sum_{k=0}^\infty 2\beta(2k+1)z^{2k}, \\ &\qquad\text{switch the order of summation and notice cancellation}. \end{align*}$$

We now have two series for $\frac{\pi}{2}\sec\left(\frac{\pi z}{2}\right)$,

$$\sum_{n=0}^\infty\frac{(-1)^n E_{2n}\pi^{2n+1}}{2\cdot4^n(2n)!}z^{2n}=\sum_{n=0}^\infty 2\beta(2n+1)z^{2n}.$$

And upon equating coefficients we find that

$$\beta(2n+1)=\frac{(-1)^n E_{2n}\pi^{2n+1}}{4^{n+1}(2n)!},\qquad n\geq0.$$

Answer 2

This actually $-\beta(n+1)$ (dirichlet beta function) Thanks to everyone who helped

Answer 3

$$\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{(2k+1)^{n+1}}x^k=-2^{-(n+1)} \Phi \left(-x,n+1,\frac{1}{2}\right)$$ where appears the Lerch transcendent function.

For the case where $x=1$, this reduces to $$\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{(2k+1)^{n+1}}=2^{-2(n+1)} \left(\zeta \left(n+1,\frac{3}{4}\right)-\zeta \left(n+1,\frac{1}{4}\right)\right)$$ as given in comments.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\sum_{k = 0}^{\infty}{\pars{-1}^{k + 1} \over \pars{2k + 1}^{n + 1}}\,\right\vert_{\ n\ \in\ \mathbb{N}}} \\[5mm] = & \sum_{k = 0}^{\infty}{1 \over \bracks{2\pars{2k + 1} + 1}^{\,n + 1}} + \sum_{k = 0}^{\infty}{-1 \over \bracks{2\pars{2k + 2} + 1}^{\,n + 1}} \\[5mm] = & {1 \over 4^{n + 1}}\sum_{k = 0}^{\infty} {1 \over \pars{k + 3/4}^{\,n + 1}} - {1 \over 4^{n + 1}}\sum_{k = 0}^{\infty} {1 \over \pars{k + 5/4}^{\,n + 1}} \\[5mm] = & \bbx{{1 \over 2^{2n + 2}}\bracks{\zeta\pars{n + 1,{3 \over 4}} - \zeta\pars{n + 1,{5 \over 4}}}} \\ & \end{align} where $\ds{\zeta\pars{s,\alpha}}$ is the Hurwitz Zeta Functon.