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Problem #5: Determine which of the following series converge 7_2 2 "tSn 2 + 6n + 3 (iii) Vnll + Sn 2 n' +n(A)and (iii) only (B) and (iii) only (C) and (ii...

Question

Problem #5: Determine which of the following series converge 7_2 2 "tSn 2 + 6n + 3 (iii) Vnll + Sn 2 n' +n(A)and (iii) only (B) and (iii) only (C) and (ii) only (D) none of them (H) all of themonly(iii) only (F)onlyProblem #5:SelectJust SaveSubmit Problem #5 for GradingProblem #5 Attempt #1 Your Answer: Your Mark:Attempt #2Attempt #3Problem #6: Which of the following series converge?2 7n+1 2 (iii) arctan(n)none of them (B)only (C) all of them ()and (iii) onlyand (iii) onlyand (ii) only

Problem #5: Determine which of the following series converge 7_2 2 "tSn 2 + 6n + 3 (iii) Vnll + Sn 2 n' +n (A) and (iii) only (B) and (iii) only (C) and (ii) only (D) none of them (H) all of them only (iii) only (F) only Problem #5: Select Just Save Submit Problem #5 for Grading Problem #5 Attempt #1 Your Answer: Your Mark: Attempt #2 Attempt #3 Problem #6: Which of the following series converge? 2 7n+1 2 (iii) arctan(n) none of them (B) only (C) all of them () and (iii) only and (iii) only and (ii) only (iii) only (H) only



Answers

Which of the following geometric series converge? (I) $20-10+5-2.5+\cdots$ (II) $1-1.1+1.21-1.331+\cdots$ (III) $1+1.1+1.21+1.331+\cdots$ $(\mathrm{IV}) \quad 1+y^{2}+y^{4}+y^{6}+\cdots,$ for $-1<y<1$ (a) (I) only (b) (IV) only (c) (I) and (IV) (d) (II) and (IV) (e) None of the other choices is correct.

Take a look at this series here, we're trying to see which one of these are true for So whether or not it's a divergent conversion to are converging. Absolutely. So let's take a look at this by the root test here. It's going to take the in fruit of this whole thing Sine of X to the end power air over to to the and power and and squared here. Taking the interest of all that here we get sine of X on top, we get to on the bottom here and then we have an end to the two of her in power element has that approaches infinity of that. So the fact that we have to over end power here that goes to zeros and goes to infinity and then anything to use your power is one. This will approximate to absa value of sine of X. Over to the nice thing about this, is that well after value of sine of X is between zero and one, then this is going to be less than Or equal to 1/2. Just definitely listed one. And so because of that here, can see that it's absolutely convergent and thus it's also accomplishment there. And so this is choice. So basically it would be the one, the choice that has Roman numeral two and 3 for the convergence of this series. An absolute convergence of it. And that's your answer there

In discussion. We have the cities one plus one upon 3 plus one upon 9 plus one upon 27 plus so on. We need to find the correct number four which the serious converges and the options are available. So let's see how does all discussions first of all, let's find the ratio of second term to first term and this will be calls to one upon three divided by one. Hence ratio will be close to one upon three. Similarly, when we find the ratio of 3rd time to second term, we get one upon 9 divided by one upon three. So this will be calls to one up on three. So we can observe that every time we are getting the same ratio. Hence we can conclude that the given series is a geometric series with common ratio R equals to one upon 3. And from the cities we can see that the first two term Of the city's is even if equals to one. We know that for a series to convert the common ratio must be less than one And one x 3 is less than one. Hence the series converges. And now let's find the some of the cities the formula to calculate the sum of. In final dramatic cities can greatness as he recalls to first term even divided by one minus commendation are so now substitute all the values we get summit recalls to one upon 1 -1 upon three hand, Some will be close to three x 2. Therefore we can conclude that The series Converges 4, 3 x two. Thus option C. Is the correct answer. I hope you wonder for the solution. Thank you.

This is a continuation of the previous exercise. So they tell us to use the result from this. So if you haven't actually solved 61 yet, I would go ahead and do that purse before you attempt this one. But with that result. So what we showed was this interval Onley converges when P is strictly larger than what. And then we came to the conclusion that this Siri's over here will also converge on Lee if P is larger than one. So all we really need to do is get the Siri's into this form here. And once we do that, we can say whether they converge or diverge. So this 1st 1 here is really with P is equal to one. So that's going to imply it. Die purchase. And this next one we have p is even toe 1.1. So actually, I should probably be doing this. This is must then or equal to one so diverges. So this is gonna be greater than or equal to one which implies it converges now or these two here. Well, we need to rewrite this integral or this serious first because it's not quite in the form of what we have. So we can pull that three out from being the power like this, and then we can just pull it out of our Siri's entirely. So we have this here and now we could go ahead and rewrite this us this being to the first power. And so just like for part A, this implies it diverges sex. P is even toe one which is less center equal to one. And then lastly, for this one, this is where P is equal to three. So that is strictly greater than one. So that implies this con urges. When I just realized over here, I need to say that this is strictly greater than one as opposed to equal to. But this is how we can use that fact that we showed in the previous problem that each of these converges or diverges

Does he form? Joint of three. Austria eight Blessed for under 50 plus five out of 20. Far less six out of 35. And so arm here we can rid rent us. Now, as the day goes from one to infinity, they would have a K plus one. Do you writing by And, uh, here we have the K uh, one square minus one. And then we can write this down. Incident is a way here. So I would start him to Western from one here. I don't have the gay and then K square minus one. Here. We noticed that. And, uh okay, square minus one. It will be smaller than the case square and every fluent over weekend in one of the case. Grandmothers, When it's bigger than the one of a case square on that may lead one of everything. Great guy, we will have excel. It is when you leave that term here and now it isn't. It can be signified. You're one of a day. On the other hand, we know that some simple one of okay this is here will be divergent by the harmonics trees. Therefore, with this and this cynical inequality here we conclude that this race he will be divergent by the Comm person past


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