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?

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.
