Answer
$\frac{2^{p}}{2^{p}-2}$
Work Step by Step
$$\frac{(1+\frac{1}{3^{p}}+\frac{1}{5^{p}}+\ldots)+(\frac{1}{2^{p}}+\frac{1}{4^{p}}+\ldots)}{(1+\frac{1}{3^{p}}+\frac{1}{5^{p}}+\ldots)-(\frac{1}{2^{p}}+\frac{1}{4^{p}}+\ldots)}=\frac{S+T}{S-T}$$
$$2^{p}T=1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\frac{1}{4^{p}}+\frac{1}{5^{p}}+\ldots$$
$$2^{p}T=S+T$$
$$2^{p}T-2T=S-T$$
$$T(2^{p}-2)=S-T$$
so:
$$\frac{S+T}{S-T}=\frac{2^{p}T}{T(2^{p}-2)}=\frac{2^{p}}{2^{p}-2}$$, $p \gt 1$
