Answer
$$
\frac{1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\cdots}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots}=\frac{1}{(1-2^{1-p})}
$$
Work Step by Step
Suppose that the given expression be called Z.
$$
Z=\frac{1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\cdots}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots}
$$
Then
$$
\begin{aligned} Z &=\frac{1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\cdots}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots}=\\
&=\frac{1+\left(2 \cdot \frac{1}{2^{p}}-\frac{1}{2^{p}}\right)+\frac{1}{3^{p}}+\left(2 \cdot \frac{1}{4^{p}}-\frac{1}{4^{p}}\right)+\cdots}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots} \\
&= \frac{\left(1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots\right)+\left(2 \cdot \frac{1}{2^{p}}+2 \cdot \frac{1}{4^{p}}+2 \cdot \frac{1}{6^{p}}+\cdots\right)}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots} \\
&=1+\frac{2\left(\frac{1}{2^{p}}+\frac{1}{4^{p}}+\frac{1}{6^{p}}+\frac{1}{8^{p}}+\cdots\right)}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots} \\
&=1+\frac{\frac{1}{2^{p-1}}\left(1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\cdots\right)}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots} \\
&=1+2^{1-p} Z \end{aligned}
$$
Therefore,
$$
Z=1+2^{1-p}Z
$$
$\Rightarrow $
$$
Z-2^{1-p}Z=1 \Rightarrow Z(1-2^{1-p})=1
$$
$\Rightarrow $
$$
Z=\frac{1}{(1-2^{1-p})}
$$
So,
$$
\frac{1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\cdots}{1-\frac{1}{2^{p}}+\frac{1}{3^{p}}-\frac{1}{4^{p}}+\cdots}=\frac{1}{(1-2^{1-p})}
$$
