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# Determine whether the sequence converges or diverges. If it converges, find the limit.$a_n = 2^{-n} \cos n \pi$

## $$\therefore \text { by Squeeze Theorem, } \lim _{n \rightarrow \infty} \frac{\cos n \pi}{2^{n}}=0$$

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what makes the limit as n goes to infinity of a m. A little bit off road Is this co sign of empire? Because Lim has in goes to infinity a co sign of N pi. That limit does not exist. Okay, so coastline of empires the thing that's causing us some trouble here, But we can still work with it. Co sign of empire is bounded. There's going to be somewhere between minus one on one. So regardless of what Innes co sign of empire is going to be between minus one in one. And now if you multiply by two to the minus stand, which is the same thing as divided by two to the end, then we still have these bounds here because Chechnyan is just going to be some positive number. So we're multiplying or divided by a positive number. Then we don't have to worry about switching these inequality signs and then we should start with the same thing. Even if we do limit as n goes to infinity, this is something that should be trooper all in. It's on the far left as n goes to infinity minus one over to the end is going to go to zero in the middle by construction. This should be limited in goes to infinity A n So you should have it worked out where you get in there and then on the far right as in goes to infinity one over to win is going to go to zero as well. So now this guy is screams between zero and zero. So the only logical conclusion is that women as n goes to infinity in a M is equal to zero. So I sequenced does converge and it converges to zero.

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