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Problem 52 Hard Difficulty

Determine whether the sequence converges or diverges. If it converges, find the limit.
$ a_n = n - \sqrt {n + 1} \sqrt {n + 3} $

Answer

converges to $-2$

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Video Transcript

for this problem, It might help the rewrite. This as in minus square, Rita imprinted sees in plus one times in plus three. And I said that it might help to write it like this, because once you just have one square root function happened here, it might remind you that, you know, we could we have this trick where we've multiplied by the top and bottom by the congregate. If we're we're trying to figure out what this limit is and that's that's exactly that trick that will want to employ. Here we take our expression. All right, think about it as a fraction. So just divide it by one, and then we multiply the top in the bottom by the conjugation of of this guy. Okay, So we can just leave the inn alone and then change the sign from negative to positive here. And certainly if we want this to be inequality than if we multiply by something, we have to divide by the same thing. Okay? And now wants you work all this out. The top part we have in times. And so yet we haven't and squared. We have in times the square root thing and then we also have the minus square root thing times. And so the next term's air going to cancel out and we have this in squared minus in, plus one times in plus three. That's just That's a typical thing that happened when you multiply by the conjure git, you just get this thing squared times this thing squared. Now we can Don't forget that we're still dividing by this thing over here, okay? And now it might help if we distribute this stuff that we have over here. So this is in squared minus. And now if you want to distribute those, we have in times ends that gives us an and squared we have in times three. We all said in terms one So plus foreign. And then we have one times three. Thanks for the He's in squared. They're going to cancel. And that we have minus for in. And this minus sign remember, this minus sign distributes is well, so we have minus foreign minus three, divided by in plus square root of in plus one times in plus three. Andre would do the standard trick where we look in the denominator and we think about what term is going to infinity the fastest. And then we divide the top in the bottom, by whatever that term. Maybe so here it looks like we have ah, you know, an end times in term happening. But remember, we're taking the square root of n plus one times in plus three. So in the long run, this guy here really just grows at a rate at the same rate of in right square root of n times in is just in. So really, the thing that goes to infinity, the fastest and the denominator is just in. So we're going to just want to divide the top in the bottom by regular in. Okay, so whatever you think is going to infinity the fastest and the denominator divided the top on the bottom by that. So now we divide the top by and we get minus four minus three over in, and then we divide the bottom by end, we get one plus. And then if we want to pull the inside of the square root, then we can write it as square root of in plus one Arms and class three, divided by in squared, right because we wanted to divide the thing under the radical by N. And if we want to pull the inn inside of the radical, then we have to divide by in squared. Okay, so this is just algebra from here. Limited n goes to infinity of minus four, minus three over in, divided by one plus square root of So let's remind ourselves what in plus one time in plus three is? Okay, we said in class one times industry is in squared plus foreign plus three. So now when we take lemonades and goes to infinity, it becomes more clear exactly what's going to cancel out. So this n squared, divided by N squared is one and then we have plus four over in plus three over and squared. And now this limit has started to become pretty clear because we know that this goes to zero. You know that this goes to zero. We know that this goes to zero. So we're just left with minus four divided by one plus square root of one so minus four, divided by two, which is minus two. Okay, so we converge and we converged to negative too. So I know that this was, ah, long problem. So let me just summarize everything. So what we did was we wrote this in a different ways that we could have everything under one square root. And once we wrote it like that, then it was more clear that we're supposed to use this multiplying by the congregate trick. And then once we multiplied by the congregate, then we just had a look in the denominator and try and think about what was growing to infinity. The fastest looked like, ah, growth of in was going to be the fastest growth that we were experiencing and the denominator. So then we just chose to divide the top and the bottom by n. And once we did that, we got over, you know, to here, no. And then instead of just dividing the thing under the square root by N, we wanted to pull the end inside of the square root so that this in squared is inside of this square root here because once we have everything under the square root, then we can, you know, make this simplification here that we made. And then once we do all of that, then it's it's clear what terms are going to go to zero, and the expression simplifies nicely