How to simplify $\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$

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.