an=2^-n cosn pi

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an=2^-n cosn pi

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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so? solve for n? or what
determine whether the sequence converges or diverges, if it converges give the limits
ohh

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\[a_n=\frac{\cos(n\pi)}{2^n}\]
naw dude
the denominator goes to zero lickety split the numerator is stuck between -1 and 1
so the answer is 0?
yes of course
oh i said it wrong the denominator goes to INFINITY lickety split, not zero
so the limit is zero
and of course that means it converges correct?
yes converges to zero
thanks sooooooo much
If you need some way to justify it, you could use Squeeze Theorem :)\[\large\rm -1\le \cos(n \pi)\le 1\]Divide each side by 2^n,\[\large\rm -\frac{1}{2^n}\le \frac{\cos(n \pi)}{2^n}\le \frac{1}{2^n}\]And let n approach infinity on the left and right most sides of this inequality,\[\large\rm -0\le \frac{\cos(n \pi)}{2^n}\le 0\]

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