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Numerade Educator



Problem 13 Medium Difficulty

Test the series for convergence or divergence.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {3^n n^2}{n!} $




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Video Transcript

for this problem Since we have factorial sze, we should be thinking about the ratio test. Typically factorial, they're kind of hard to deal with. Ah, in the ratio test they're not so hard to deal with. So these these guys here these air there a in terms the ratio test We want to see if you want to see if this is less than one strictly less than one. It's equal to one then The test is not conclusive. Okay, So like we said, these guys are r A ends. So we're looking at and plus one replaced the end in this equation with n plus one and then dividing by N is the same thing. I was multiplying by the reciprocal, so we would have in factorial here three end here and then n squared here swoops in its its a fraction. Okay, so then ah, in orderto look to see what cancels out. We don't want to look at you know this three This three So here we have three to the n plus one. Here we have three to the end. You want to be comparing those guys? See what cancels out? It's all of the copies of three you're going to cancel out, except for one copy of three which will be left over from up here. Get one copy of three remaining up top in factorial divided by N plus one factorial. We're going to get a lot of things to cancel out, except for in plus one. Down here to remember in Factorial is just one times two times, three times for dot that all the way up to times in. So if you just write out the definition of what the factorial function is, then you'LL see that there's going to be lots and lots of cancellations. Everything will cancel except for in plus one down here. So then what we're left with is and plus one square divided by n squared So we can write that as in plus one over n squared can. Then if you want, you can. If if this isn't clear, what this limit as you can split it up like this. Okay, so this is something that we're allowed to do as long as we don't split it up and something that puts us an indeterminate form like, you know, if this limit was infinite and this limit zero. Then we wouldn't be allowed to do this type of thing. But in this case, this limit is going to be zero. And this woman is going to be something finite. So zero temps finite is not in determining the form. So we're definitely allowed tow split up the limit in this way and this limit If you just pull this limit inside of the parentheses, you'LL see that this is just going to be, uh, one. So then we have one squared, which is one you get, like, zero times one, which is zero. And again, the important part there is that it's something that's less than one. So we do get convergence, okay? And if this is obvious enough that this limit is zero, you don't have to split it up in the these two limits like this. But it's a nice intermediate step. If you're unclear about this, another approach you could do is you could have You could write that you know this. Copy it and plus one gets cancelled out with one of the copies have been plus one over here, right? So you don't think it was something like that and then you could use local towers rule. There's somethingto Figure out that this goes to zero or you could do the trick. You know, once you ended up here, you could use the trick where you divide both the top and the bottom by in squared and then take limit on top over limit on bottom. So there's a few different approaches that you can use here, but you should end up with this thing approaching zero and again, it doesn't need to go to zero. We just need for you to be less than one. So that zero, then we certainly accomplished that, and we get convergence.