 Finite-horizon invasion threshold of Tit for tat against unconditional defection

In an $L$-round iterated prisoner's dilemma, Tit for tat against always defect gets $S$ once and $P$ thereafter; the defector gets $T$ once and $P$ thereafter. Their self-payoffs are $R,P$. At introduced frequency $ϵ$, subtracting mixed-population payoffs gives $ϵ[(R-P)-(T+S-2P)/L]-(P-S)/L$. The displayed positive threshold applies when its denominator is positive and the threshold is less than one. Taking $L\to \infty$ first removes this finite entry cost, so the horizon and rare-mutant limits cannot be casually interchanged.