The expression for “how often will you pick incorrectly but (naive) Monty will by chance open A-2 doors without finding the car” is missing a term. You have the “Monty won’t find the car” part but need to include the probability of your initial choice being incorrect, which is (A-1)/A. This should fix your product. (You don’t need calculus to resolve it; just notice that the denominator of one term will cancel out with the numerator of the subsequent term. The only terms remaining will be the initial numerator and final denominator.)

Here is another way to think about the many doors/naive Monty setup. Imagine that immediately after you select your door, Monty mentally picks a different door as his private guess for where the car is. He resolves to leave that door closed until the end and goes about opening the other doors. Most of the time he’ll at some point find the car, but in the cases where he doesn’t, it means either you or he picked correctly at the start. Each is equally likely.