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QuestDetermine whether the following series converggs. Justify your ansurer i6+4' Select Ihe correct choice below and, necessary; fIl in the answer box lo comp...

Question

QuestDetermine whether the following series converggs. Justify your ansurer i6+4' Select Ihe correct choice below and, necessary; fIl in the answer box lo complete your choice (Type an exact answer ) The limit of the tetrns of the series tho series diverges by the Diveigerce Test The series geometric geries writh comnion ralio This less thon 80 the series converges by Ihe praperties of _ geometric series. The limit of Ihe lermg of the series is so the series converges by Ihe Divergence Test

Quest Determine whether the following series converggs. Justify your ansurer i6+4' Select Ihe correct choice below and, necessary; fIl in the answer box lo complete your choice (Type an exact answer ) The limit of the tetrns of the series tho series diverges by the Diveigerce Test The series geometric geries writh comnion ralio This less thon 80 the series converges by Ihe praperties of _ geometric series. The limit of Ihe lermg of the series is so the series converges by Ihe Divergence Test The seres geometric seric? wlth common ralio Thls greater Ihaq thc serles diverges by the propertios of a geomelric series The limit of Ihe terms of the series does not exist; 50 the serie; divarges by Ihe Divergence Tost



Answers

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$\sum_{n=1}^{\infty} \frac{-2}{n \sqrt{n}}$$

We want to determine whether this convergence or divergence and give a reason as to why. So First, let's go and just pull that five out. And I'm also going to ship that so we can rewrite this as Jay is equal to in plus One. So if that's the case, then our new lower bound is going to be J is equal to two over, bound, still infinity. And then over here, we're going to have just one over J. Well, this is the harmonic Siri's harmonic series, and we know the harmonic series Die Vergis. So our original function is really just the shifted harmonic series. So if this one diver just so must our original

We want to determine whether this series converges over diverges. Well, first any time we have something whereas like Factorial czar like powers of N, um, kind of set up like this normally is a good idea to go ahead and use the ratio test. Let's go ahead and test that out to start. So but people were look at a n plus one n a in this is going to give us. So it was gonna be five n plus one over or in plus one plus three. And then do we divide by A In what? We just reciprocate that So it b a n plus three over five in. Now we can simplify this down to be so would have five the new mayor and then foreign plus three uber foreign plus one plus three. Now, at this point, what we could do is try to take limit, as in goes to infinity and something that might help us before we actually do. That is if we divide each the numerator and the denominator by the largest power, which is going to be this foreign plus one, So that's going to give us so 1/4 plus 3/4 and plus one. And then this is going to be all over. So that would just be one plus 3/4 and plus one. So we have this here now, if we were to take the limit. So we're supposed to cops value, but this is always gonna be positive, so no issues with that. But now, if we were to go ahead and take the limit of this as in approaches infinity, Well, this is gonna go to zero. This is going to go to zero. And so we're going to get that the limit. L is equal to just five force, and this is strictly larger than zero. A trickle larger than what? And so that implies that this will diverge by the ratio test.

We want to determine whether the Siri's coverages or diverges now I think this would be a good candidate or interval tests cause the hyperbolic functions. I really don't know how to deal with, like for ratio, test or root tests or anything like that. So let's go in and check the criteria. So if we let this be ableto f well, we know what want to Infinity Central been is going to be positive. And if we square that, that's still gonna be positive. So F is going to be strictly great at zero for in Greater Than You go to one. So no, she's there. We also know such is also defined for all values. So it's gonna be continuous cause. Remember, it's one of her kosh and cautious, always going to be positive. And so now we just need to look at the derivative so f primes. We're gonna you do two sec and then we have to take the derivative sec, which is supposed to be negative sec of 10th or negative times tent. And this here is going to be less than zero or in greater than equal one, because since squared in tow, 12 in February 7 Deposit and tents of In is going to be positive. All want to impending as well. So all the criteria rentable testament. So let's go ahead and integrate this. So we're gonna integrate from one to infinity. Oh, sick squared. And this is just 10th. So we evaluate this from one to infinity, and the limit, as Inglis would've been need for 10th goes to one so you could go ahead and do one minus tense of one. And I have no idea what that value is by really Don't Carol that much, because the only thing that I care about is that this convert so are integral convergence. So that implies, by the integral test that are Siri's converges as well. So applies, and is he gonna want affinity? Stick Xa n con verges to

We want to determine whether this Siri's converges or die Burgess. So the first thing I would do is try to apply the ratio tests to this, because any time we have where it kind of looks like a Paul no meal over a polynomial, it's normally a good idea to go about using that. So let's look at a seven plus one over a even first that we comply. The limited everything later, but right now this is all we really care about. So that's going to be e to the end, plus 1/1 plus e to the two n and plus one. And actually, let me just write that out as to end plus two and then we're gonna multiply by. It should be the reciprocal of what we had over here. So one plus e to the two in over e to then and this is a multiplication. Now we can go ahead and simplify this down a little bit. So the e to the M plus one over eat it and just becomes he. And now to simplify this, what I'm gonna do is I'm going to multiply the top and bottom by the largest power. So there's E to the two and plus two. And so it may not be obvious why I want to do that now, but it will be in one room. So we multiply by one direction. All right. To says e to the negative, to end minus two over E to the negative two and minus two. So simplifying all that down, So we still have that e So e times e to the negative two and minus two. And then we have plus e to the negative, too. And then all over e to the negative two and minus to plus right now, if we were to take the limit, as in approaches infinity of this, this is going to give us so e times easily negative. Second all over one which, if we simplify that down, would give us one over e. So this is strictly less than one. And so this implies by ratio test. It will con virtue


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