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Problem 56 Hard Difficulty

Determine whether the sequence converges or diverges. If it converges, find the limit.
$ a_n = \frac {(-3)^n}{n!} $

Answer

$a_{n}=\frac{(-3)^{n}}{n !}$ converges to 0

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Video Transcript

if this problem it looks, looks like limit, as in goes to infinity of am Looks like we should be getting zero here. So I said that because fact where growth is faster than exponential growth, okay. And sometimes working with the absolute value of function is easier than just working with the original function. And if it is true that this limited zero, then we would want to consider that option. So if we can show that limit as n goes to infinity of the absolute value of a N zero, then certainly when as n goes to infinity of a N is going to be zero, right? If absolute value of a and a zero, then the only way that can happen is if a and itself goes to zero. Okay? And if it doesn't, then this approach isn't going to be so useful. So you like to start with the intuition and realized that pectoral growth is faster than exponential growth. So this is what we expect to happen, okay. And then we just start working with the absolute value. Okay, So when we're taking the absolute value instead of having minus three to the end, we'LL just have through to the end, divided by in factorial and another way to write this What's so? I guess there should be There should be a limit here. So I'll just write it on the next page limit as in goes to infinity of three to the end, divided by in factorial And then another with a right with is limit as n goes to infinity of three times three times three times that That that times three where we have a total of in copies of three there and then we're dividing by one times two times three times dot that thought times pin Okay, So again we rewrite things I have three ever won number two three over three whatever for? And then we go all the wayto three over in. Okay, so for your place, all of these terms with three over four, then we should get something that's larger or at least equal, right? Because think about, you know, through over and through over five. Bob a block, all the stuff in between. All those terms are going to be smaller than three or four. So if we replace them with three over four, then we should get something bigger or at least equal says it will do here. We're just making some replacements. So we're just saying replace all these terms with three over for and that should give us something bigger or at least equal. It's now we're just gonna have some power of three over four here. So we think about, you know, the fact that we started with in terms and then we've used up through about terms over here. So now we have in minus three terms left. Okay. And then, since no absolute value of three over four is less than one we have, we have that the limit as n goes to infinity through over four to the end, minus three. That has to be zero, right? Do you have a number that's less than one absolute value? And you're looking at the limit as the exponent goes to infinity. Then you get zero. So if in goes to infinity certainly in minus three is going to go to infinity as well. Okay. And if this happens, then we just replace this guy here with zero. And we have that limit as in goes to infinity of absolute value of a N. This has to be less than or equal to three over. One times three over two times three over three times zero and multiplied by zero just kills everything. Okay, so this limit is less than or equal to zero. But we're working with absolute values. So absolute values, everything should always be positive, or at least non negative. Okay, So if everything is non negative and we're less than or equal to zero, the only way that this can happen is it. This limit is equal to zero. Okay. And if this limit is zero, then the only way to have something that is zero in absolute value is if the thing you're working with is of course, zero as well. So therefore, limit as n goes to infinity of a n zero. So the sequence converges. Can it converges to zero