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Test for the convergence of the series -1)"5" n=0 4"...

Question

Test for the convergence of the series -1)"5" n=0 4"

Test for the convergence of the series -1)"5" n=0 4"



Answers

Using the Direct Comparison Test In Exercises $5-16,$ use the Direct Comparison Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^{\infty} \frac{1}{4 \sqrt[3]{n}-1}$$

Hello. So here we are going to be using the limit comparison test where we have a seven is equal to foreign cubed plus three N over. N. To the fifth minus four N squared plus one. So we're going to compare this with Bc. Even where we have the B seven is going to be equal to end cube Over and to the 5th. Um which is going to be equal to well just one over and squared. Now for the limit comparison test we take the limit As N goes to infinity of ace event over peace event. So that is equal to four and to the 5th plus three and cubed over and to the fifth minus four and squared plus one. Well as N goes to infinity this limit is going to be equal to just the ratio of course leading coefficients which is gonna be equal to four and we have that well zero Is less than four which is less than infinity. So therefore the series are going to converge or diverge together. Now be seven um which is while one over and squared is going to be a P. Series, with P. E. Being equal to two. So therefore be cement converges. Therefore our given series can verges as well. Okay.

Give it called zero to Infinity for to the end, divided by five to the end Class three And we would like to apply direct comparison to find a limit. So the first observation is it would be nice if we only had this because this we could write us for fifth to the end, and then that would make that would make a geometric serious and then we would know the answer. So then we can compare with that. And then how do we compare How do we compare? So let me write it down here. How do we compare four to the end over 5,000,000 plus three. How do we compare it with just for to the end, over five to the end? Well, it turns out we're lucky this is actually less, uh, for equal than that and simply because three is a positive number. So we're adding apart if we have an extra positive member into the nominator that makes this fraction smaller than this one. And this is smaller than this than the serious is smaller than this series. And this one we can write us some equals your to infinity of we can distribute that and the power. This is a geometric serious. We know exactly. We know exactly the answer, but it's enough to say since sends 4/5 you know it's in the interval. Negative 11 that means some 4/5 to the end. Convergence. And then coming back to this because this serious is smaller are serious, is a serious made out of positive terms, and it's smaller than another serious, which converges. Therefore, this one completely is.

For the theories of five and oh, over into the fourth and the series of Infinity two and equals one. We're going to be taking the route test. So the first step is to put it into the format. And the format is very important because sometimes as the problem goes on, you will forget it and you get knocked out for points. So we will have the limit as n approaches Infinity Absolute value bar. Oh, and you have five and plus one all over Yeah, and plus one to the fore all over five to the end, all over and forth. So we have something that's a fraction within a fraction, and you can do whatever you like to pretty much get the fraction over a fraction undone. So this is what I call part to simplify. What I'm going to do is it's called the Butterfly Method, something that's common commonly actually taught in algebra courses. But you will end up with five and all over and plus one to the fourth times and to the forest all over five and plus one. Uh huh. You pretty much want to make sure you're basis of five and your basis of n are across from each other. Now, if we write out our more expanded 15 to the young power times five to the first power for looking at this, then the five to the end of power is cost out. But we will have a five left over endless one to the four power. That's a lot of factoring, but if you do factor it out completely, you will have a end to the fourth was for and cubed the six and squared plus four end last one. As you can see the end of the forest cross up now. What you have is we're going to three. Take the limit so you have a limit as n approaches infinity. What's the value bar of one all over? You will have to five times four and cute plus six n squared plus four and plus money. And I put it this all in parentheses because it's from the characterization of Empress one to the fore. Make sure you close that absolute value bar. Now, Our next step of this is actually to take out any coefficients that can be taken outside of the limit. And this case is the fifth, and then you get left with the rest of the limit. You just make sure you write out the rest of the limit because you can get docked down for points if you don't. So we just have everything left. Now we're going to do is solve it and every plug in n two. What's glad? So pretty much plucking an infinity. We will have a fit times one all over infinity. Practically This within itself gets canceled out. So we will have. Fifth is our answer Now. Fit is less than one, which leads me to set for Is it a convergent divergent? You know, as a absolutely convergent it's an equal to one. So is it and decisive? What's the answer? So we know here that convergent uh, it's awesome. One divergent is greater than one. Absolutely conversion. There's actually it would also be greater than one so convergent and absolutely convergent are kind of the same thing here in the root test and in this case and conclusive would be equal to one, and suddenly, we can say, is less than one. Her final answer of the series, Infinity and equals one five to the power all over and to the fourth is conversion. Hope you understand more on the route test, and that's how you solve that problem.

We should move Calling about the limit comparison Danced would send that if we have the limit on the PNR, I am and goes to infinity coaching some constancy policy between the desert to infinity That we will have the behavior of the I m submission will be similar to the behavior of the submission of the PM in this question. Were given the submission under and broke a minus one, if any by the end. Okay, that's one from one to infinity. Okay, here will be created into and now we have given this one will be the i n. We need to choose the B and now so my chick here will be I think a maximum bar on the top will be invoke a minus one. The maximum bell on the bottom will be end. Okay, It was simply find this one to get me go to one of And on the other hand, we know that the submission of one of an there's an end is a harmonic Siri's. Therefore, we should get this one go be divergent and the next time we need to compute the limit on the b end of i m so one of the N times the reciprocal will be the end. Okay. Plus one divided by N about K minus one that we should get equal to the limit on the and okay, just one of a number. And okay, so you get a call to the limit off the one plus one of the end. Okay, So when n goes to infinity, this one goes to the zero. Then again, the limited which one? And it's greater than zero on now by the limited person tested with this limit and this divergent here, this raise him was also divergent.


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