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# Determine whether the sequence converges or diverges. If it converges, find the limit.$a_n = \frac {(-1)^n}{2 \sqrt n}$

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limit as n goes to infinity of a n This is limit as n goes to infinity of minus one to the end divided by two square root of end Okay, Miss Lim is going to be zero. So if this minus one to the end is throwing you off you need convincing of this fact consider limiters and goes to infinity of absolute value of a And certainly, if this goes to zero then we also must have that this goes to zero. The only way for a and to go to Zuma would be in absolute value of a and went to zero and vice versa. Okay, so if you need convincing of this, just consider this and it should be should be clear Susan goes to infinity. If we're just looking at absolute value, we just have one over to the square root of n Okay, this is one is just some finite Number two is just some finite numbers of these. This one and two aren't really going to do much. And then the square root of n squared of n is definitely going to go to infinity as in goes to infinity. So this is going to look like one over infinity. Which is of course, zero. Okay, so this limit is zero, which means that we we converge two zero. And again, I was just doing the absolute value here. Tto convince you that this minus one to the end and our original expression wasn't going tto mess up The fact that this limit does indeed go to zero.

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