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9.(22 pts) Determine whether the series is convergent or divergent: You must support your answer with valid reason for answering as you did. You need to use the fac...

Question

9.(22 pts) Determine whether the series is convergent or divergent: You must support your answer with valid reason for answering as you did. You need to use the facts and theorems that were established in class andor in sections 1L.l 112.14.3, and 11.4 to support your answers In these problems you do not have to find the sum of any convergent seriessin(nx) Vn

9.(22 pts) Determine whether the series is convergent or divergent: You must support your answer with valid reason for answering as you did. You need to use the facts and theorems that were established in class andor in sections 1L.l 112.14.3, and 11.4 to support your answers In these problems you do not have to find the sum of any convergent series sin(nx) Vn



Answers

$21-34$ Determine whether the series is convergent or divergent.
If it is convergent, find its sum.
$$\sum_{n=1}^{\infty} \sqrt[n]{2}$$

For this problem, we are asked to determine whether the sum from n equals one to infinity of 0.8 to the power of n minus one minus 0.3 is convergent or divergent. And if it is convergent we are to find it some. So to begin, one thing that we can do is recognize that we are actually essentially dealing with two different sums here. We have the sum from n equals one to infinity Of 0.8 to the power of N -1. And then we are subtracting off the sum from n equals one to infinity Of 0.3 to the power of N. Which itself we can write as 0.3 times 0.3 to the power of N -1. Reason why I say that is because you can see that our first series there, we can write as a geometric series with a equals one, r equals 0.8. We know that our our is less than one. Therefore that first term is convergent. Okay, And our second term Is a geometric series with a equals 0.3 and r equals 0.3. So we can see again are less than or not are less than zero are less than one. Which means that we have convergence as well. So we can apply our geometric series formula for this. So we'll have a over one minus are so 1/1 minus 0.8 minus 0.3 Over 1 -0.3 Calculating that all out the final result is going to be 4.57145 Approximately, or 143 rather approximately. Yeah.

We want to determine whether this series converges or die. Vergis. Now, whenever we have something where it looks like a Paul no meal over a polynomial, it's normally a good idea to check to see. Um, if we can apply ratio test, So let's go ahead and do that. And actually, before we do that, I'm gonna go ahead expand. What we have in the denominator is gonna b e to the end over e squared in plus 20. Eat in plus 100. Now, let's go imply ratio test to this. So we have a n plus one over a n. So I was gonna be e to the and plus one, and that is going to be all over e to the two in plus two. I lost 20 e to the to and and then plus 100. And then this is gonna be times by so a in which I'll just go ahead and write and read here so it would be e toothy to in plus 20 e to the end, plus 100 and then more house e to the end in the denominator. Right? So if we were to go ahead and simplify this down. We should end up with the followers. So first, that just gives us E and also knows if we multiply the top and bottom by e to the negative two and minus two and I'll be able that more parent, why wouldn't do this moment? We should end up getting so on the inside here. This is gonna be e to the negative too. Plus 20 e to the negative in minus two and then plus 100 e to the negative two and minus two. And then in the denominator, that should give us one plus 20 e to the negative to and then plus 100 e to the negative two in minus two. So now if we apply the limit as n goes to infinity for this Well, this goes to zero. This goes zero. This goes to zero and this ghost zero, and we're just gonna be left with e times e to the negative second over one. And that simplifies down toe one over E, which is strictly less than one. And so this implies that this wall converge so it converges by the ratio test

For this problem, we are asked to determine whether the sum from N equals one to infinity of each. The power van divided by n squared is convergent or divergent. And if it is convergent, where to find it? Some. So the first thing that we can do here is observed that our individual terms of our series, the A. N. S. Will be E. To the power then times end squared. Now we try taking the limit as N approaches infinity of our A. N. S. We basically need to consider which is going to be increasing faster. That E to the power of N. Or that N squared. We can tell that when we exponentially eight something like doing each power of N, it grows much more rapidly than if we are just squaring something. So if we're doing each the power of N divided by N squared, well actually have that the limit as an approaches infinity is going to itself be infinity because the numerator is growing much more quickly. So we can tell then that if we're adding up an infinite number of terms which themselves are approaching infinity, then our series has to be divergent

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


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