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Numerade Educator



Problem 6 Medium Difficulty

Test the series for convergence or divergence.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {n^{2n}}{(1 + n)^{3n}} $




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Video Transcript

for this problem. The first thing that I noticed that we have an end on top and we have the one plus in at the bottom. And then we haven't in as Thie exponents. So that reminds me of this fact that this limit is one over you. Usually we see it the other way around. Ltd's in approaches infinity of n plus one over n to the power of n, is he? But certainly if you take the reciprocal of the limit, then you get the reciprocal of the answer. So this reminds me a bit of this. It would be convenient if these guys had the same exponents because then we could just lump this in over one person together and take out whatever their common exponents wass. So down here we have an explosion of three n appear we have next moment of to end. So if you want to come and exponents, what we could do is multiply the top in the bottom by and to the end. And then once we do that, then both of these guys haven't exponents of three. And now so if you want, we can pull the ECG their common exponents out and get this. Okay? And then, as we mentioned here as in goes to infinity, this chunk is just going toe be some finite constant. So in this case, this chunk is going to be won over E cute. Okay, so this is just motivation for what the being should be when we're doing the limit comparison test. Okay, so this this is going to be the in terms. So these they're just, uh terms that we started with and then the B in terms you could make one over into the end. Can't be in term. Should be somewhat close to the in terms in the sense that if we take this limit, we should get something that's finite and non zero. So in this case, as we mentioned, when we take this limit, what we're going to end up getting is one over, he cubed. Okay, so it doesn't matter that this is not equal to one doesn't really matter that it's less than one either. The important part is is that it's something that's finite, and it's non zero. And as long as we accomplish that, we know that whatever happens when we sum up these be in terms? We should get the same thing when we sum up these A in terms the same behavior that isthe so summing up these be in terms from n equals one to infinity. We're going to get convergence. So therefore we should get convergence over here. And if you really want to be convinced that summing up these be in terms, you're going to get convergence. What you could do is compare it to one over N squared, right? One over in Squared is always going to be bigger than one over into the end for in equals one all the way up to Infinity. And these are all just positive terms here. So summing up to be in terms is less than summing up the one over in squared terms, and everything's positive and one over in squared terms converge. Then these guys must converge, and therefore the A in terms must convert by the limit comparison test. But I think that it should be at least intuitively obvious that summing up be in terms will get convergence, and if it's not, then you might want to think of a little bit more about why that would be So. I mean, basically, you just want to be convinced that these terms are approaching zero sufficiently fast and as long as that's happening than you should get convergence.