Finitely repeated Prisoner’s Dilemma with switching cost I'm doing this finitely repeated Prisoner's dilemma with switching costs but I have trouble showing the fact that $\varepsilon$ had to be $1 &lt; \varepsilon &lt; 2$. I do see why and  that it is a subgame-perfect Nash equilibrium, but how do I show it.
My approach:
Consider deviations by player 1, given that player 2 sticks to his strategies in his subgames following histories that end in each of the four outcomes of the game. (By symmetry the below would also account for deviations by player 2 if player 1 sticks to his strategies).
We know that this game is played $T$ times. Let $m$ denote the number of cooperation stages (i.e. number of times the strategy profile $(C,C)$ has been played), let $n$ denote the number of punishment stages (number of times the strategy profile $(D,D)$ has been played), and let $T = m + n +1$. If the players cooperate $T$ times they get $3$ in payoffs. Thus, we can write
$$3T=3(m+n+1)$$
If player 1 deviates at any stage of the game he'll get a payoff of
$$3 \cdot m + (4 - \varepsilon) + 2 \cdot n$$
For the deviation to be profitable for player 1 the deviation payoff has to be greater than or equal to the cooperative payoff. That is,
$$3 \cdot m + (4 - \varepsilon) + 2 \cdot n \geq 3m + 3n + 3$$
We can simplify this
$$n \leq 1 - \varepsilon$$
And this is where I'm stuck. What am I supposed to interpret from this? That when $\varepsilon \in (1,2)$ then the number of punishment stages is negative?
Another question I have is: Assume that the question above got answered. Is that enough? Or do I need to show something more? Because I'm well aware that playing $(C,C)$ $T$ times is a SPNE and playing $(D,D)$ $T$ times is also a SPNE and either way there is no profitable deviations.
So, to summarize: I would like to know if I'm approaching this problem in the right way. I would like to get some help on how to move forward from where I got stuck. And if the above would be enough to show what the problem is asking for.
Many thanks in advance!
[Update]----------------
Below is my updated attempt on a solution with some inspiration from M.J. Osbourne and the feedback I got from @Herr K..
A couple hints.

Regarding the lower bound on $\epsilon$: What happens if deviation occurs at stage $T$? In other words, there is no opportunity for your so-called &quot;punishment stages&quot;.
Regarding the upper bound on $\epsilon$: Suppose player 2 deviates at stage $T-1$ but player 1 does not. What must be true about $\epsilon$ in order for player 1 to have an incentive to switch to $D$ at the very last stage?


In general, to show that the given strategy profile is subgame perfect, you need to argue that playing $C$ after $(C,C)$ and $D$ after either $(C,D),(D,C),(D,D)$ are optimal.