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Determine whether the sequence converges or diverges. If it converges, find the limit.

$ a_n = \frac { \cos^2 n}{2^n} $

converges

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for this problem will use the Squeezed Sam. So there thing that kind of gets tricky with this. A in term is this co sign squared of N to the coastline squared Event is the troublemaker here, and hopefully it's something that's bounded. And if he knows which is the case here, then you can use the squeeze there. So we bound this thing that's giving us some trouble. This's co sign squared of ends that's going to be bigger than or equal to zero and less than or equal to one. You can't be a negative number because this is co sign squared. Okay, so this is going to be something that's true for all in, and we can divide by two to the end. And it's not going to change these bounds either, because to the end is always going to be some positive. Number two, divided by a positive number, are multiplying by a positive number. We don't have to worry about flipping these things that are, and then we can put these limits on take the limit as n goes to infinity limit as n goes to infinity. No Lamma as n goes to Infinity case is on the far left. This is still zero. Now we have the limit in question here by construction. This was our way in term. And then on the right hand side, we have limited as n goes to infinity of one over to the end. So that is also going to go to zero. So this Lim is trapped between zero and zero for it squeezed between zero and zero. That's where the name the serum comes from. So the only logical conclusion is that limit, as in approaches, infinity of a N is actually equal to zero to our sequence does converge and it converges to zero.