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[email protected] where @n n=lln + 5This seriesIts sum isWhat about the sequence {an } ne_1 ?This sequenceIts limit is...

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[email protected] where @n n=lln + 5This seriesIts sum isWhat about the sequence {an } ne_1 ?This sequenceIts limit is

2n @n where @n n= lln + 5 This series Its sum is What about the sequence {an } ne_1 ? This sequence Its limit is



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Limits of Sequences If the sequence with the given nth term is convergent, find its limit. If it is divergent, explain why. $$ a_{n}=\frac{5}{n}\left(n+\frac{4}{n}\left[\frac{n(n+1)}{2}\right]\right) $$

Welcome back to numeral by name seven. Chirac. Let's consider the sequence Ace of an Equals five Times and Divide by n Plus One. And I'm going to write this equivalently, though it's gonna look different as five, divided by one plus five over end. And I've done this by dividing the top and the bottom by the largest power of the denominator. In other words, I've multiplied the top of by one over end, and I multiply the bottom two turns by one over end. So I get an equivalent expression. But it's more advantageous this way because as I consider end growing large, the five will remain unaffected. The one will remain unaffected and this five over end will go off to zero. So that'll mean that this entire thing will approach. Perhaps we should use some more proper notation here will say that this here approaches zero with the bottom right in term, going away, which would be equal than to 5/1 or just simply five. So that means that the sequence does converge. Convergence to five, and you can write that simply as being the limit as n goes to infinity of Ace of N is equal to five. I hope this video helped. If it did take a moment, click that hard at the bottom of the screen.

Even, uh, sequence under form and are endless. Five found on. And then we need to find a limit on the sequence. Yeah, energy. Do this one here. Notice that we can read. Write this down is the and plus five and bound a minus one now and then about an outside. And then I can write down just as the one plus five over and now about it. And and everything here would be better off. And then when it's one so now let trying to find a limit. I'm just sequence now. So have own outside. Here will be it about a minus one. Inside, we have the one plus five and found on the end. Now we know that we can passed the limit inside about one of about minus one. Therefore, we can again ico Jew who had a pound minus one outside. Yes, I had a limit off the one plus five. And well and and just infinity every condom that a that a about. We had, uh, able expect culture the limit on the one plus x cover and on, and it's and just infinity. So it isn't in formula here. We can identify the five pair with expel. Therefore, we should get Echo Chew. We still had about minus one outside. Inside. We have isn't recorded at a five and this one and guessed that Eva minus five. I would you would you want over able five?

Okay, so we want to find the first term's over sequence. So when N is was one, we have one plus one in one plus two all over two times 12 back to. So this is two times three, which is 6/2. So that's equal to three I know, let's find and is equal to two. So here we have two plus one and then two plus two all over two times two squared. So this is three times four so that the 12 over, um to U turns for, which is eight. Who was 12 over each. Yes, sir. This is into a to be over to, and now it's fine and is equal to three. So when N is acquitted today we have three plus one. Do you put too, all over two times? Three to depart, too. So that's four times five with just 20/9 times to which is me, too. So what's 20/18? That's equal to a 10 over nine. I know we want to find one, and it's equal to four. Well, it is four plus one and then we have four plus two on over two times foreign to depart to. That's five times six, which is 30 over two times 15 which is 32. Okay, so what's 30/32? That's equal to 15/16 and I would find and is a quantified. So we have five plus one times five close to all over two times five squared. So this is six times seven. That's 40 to over 25 times two, just 50. So this forgets you over 50. Simplifying. Oh, yes, it does this improvising to 21/25. Okay, to know that we found our first back times. Let's find her falling limit. So for our limits, what's notice? Um or let's start by multiplying through our numerator. So this gives us that we have a and squared close three un plus two all over two and squared. And now let's break this up into three fractions. So we have the limit as an approaches. Infinity of end squared over +21 squared. So that's a 1/2 plus three un over two and square instead of 3/2 1 when in plus a, um, that's going to be a born over and squared. I know taking are falling limits. This approaches their own. It's also pretty girl. So we see it. Our limit valueless one house. So our sequence here converges and it can register to 1/2.

Okay, so we're giving her following sequence, and here we want to find our 1st 5 times. So let's start with and is equal to one. So this gives us that we have 1/1 plus two. What you do to 1/3, and then we have an is equal to That's one action. It's too over two plus two. So this is equal to 2/4, which is 1/2. And now when? And as you could to drink, we have three over three plus two. So this gives us that we have 3/5. Now let's move on to on is equal to four. So this is four over four plus two. So that's for six. Before over six. This reduces into a 2/3, and I went and is able to five. We have 5/5. What do would ubiquitous seven over arching? That's five or seven. Okay. And now we want to find, um, the limits as an approaches infinity of any over and plus two. Well, let's start by dividing, um, or numerator and a denominator by any. So this gives us that we have the limits as an approaches Infinity of one over end over what? Over direction is not over. And we divided by N We get 1/1 plus two over end. So we see that this year delivered as an approaches infinity. This is equal his own, so he gets 1/1, which is equal to one.


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