Question regarding part of the proof for the typical subspace theorem

Part three (going by N&C page 544) states that
$$tr(S(n)rho^{otimes n})=tr(S(n)rho^{otimes n}P(n,epsilon))+tr(S(n)rho^{otimes n}(I-P(n,epsilon))).$$
Now I understand how the term on the left of + goes to 0 as n $to infty$. However, I am confused how the term on the right does. N&I states that you can set
$$0 le tr(S(n)rho^{otimes n}(I-P(n,epsilon))) le tr(rho^{otimes n}(I-P(n,epsilon))) rightarrow 0,,text{ as } nto infty.$$

I don’t quite understand why this is the case. My only assumption is that the eigenvalues of $rho^{otimes n}(I-P(n,epsilon))$ are bounded in such a way that as $n to infty$ it will go to zero. However, I am unsure how to go about calculating this bound, though I assume it is of a similar form to the eigenvalues of $rho^{otimes n}P(n,epsilon)), 2^{-n(S(rho)-epsilon)}$