How to simplify $\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$ Let $p$ is real number which satisfies$\quad p > 1$
How can I simplify the fraction
$$\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$$
Numerator is $\zeta(p)$ but I don't know the closed form of denominator.
Is there any idea to simplify this fraction?
 A: Hint. We have $p>1$. Using the absolute convergence of the series, one may consider
$$
\left(1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots\right)-2\left(\frac{1}{2^p} + \frac{1}{4^p} + \frac{1}{6^p} + \cdots\right)=1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots
$$Then one may write
$$
\frac{1}{2^p} + \frac{1}{4^p} + \frac{1}{6^p} + \cdots=\frac1{2^p}\left(1+ \frac{1}{2^p} + \frac{1}{3^p} + \cdots\right)=\frac1{2^p}\cdot \zeta(p).
$$
Hope you can take it from here.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{{1 + 1/2^{p} + 1/3^{p} + 1/4^{p} + \cdots \over
1 - 1/2^{p} + 1/3^{p} - 1/4^{p} + \cdots}} =
{\zeta\pars{p} \over
\sum_{{\large n = 1} \atop {\large n\ \mrm{odd}}}^{\infty}1/n^{p} -
\sum_{{\large n = 1} \atop {\large n\ \mrm{even}}}^{\infty}1/n^{p}}
\\[5mm] = &\
{\zeta\pars{p} \over
\sum_{n = 1}^{\infty}1/n^{p} -
2\sum_{n = 1}^{\infty}1/\pars{2n}^{p}} =
{\zeta\pars{p} \over
\pars{1 - 2^{1 - p}}\sum_{n = 1}^{\infty}1/n^{p}}
\\[5mm] = &\
\bbx{1 \over 1 - 2^{1 - p}}
\end{align}
A: You might want to try 
\begin{align*}
\frac{1+\frac1{2^p}+\frac1{3^p}+\frac1{4^p}+\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}&=\frac{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}+2\cdot\frac{\frac1{2^p}+\frac1{4^p}+\frac1{6^p}\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}\\
&=1+2\cdot\frac{\frac1{2^p}+\frac1{4^p}+\frac1{6^p}\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}\\
&=\ldots
\end{align*}
