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5 Use the integral test to determine the convergence or divergence of the series ne You may assume n=u that function f (x) = xe satisfies the conditions necessary t...

Question

5 Use the integral test to determine the convergence or divergence of the series ne You may assume n=u that function f (x) = xe satisfies the conditions necessary to apply the integral test.

5 Use the integral test to determine the convergence or divergence of the series ne You may assume n=u that function f (x) = xe satisfies the conditions necessary to apply the integral test.



Answers

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. $$\sum_{n=1}^{\infty} \frac{2}{3 n+5}$$

Pushing Marikana Bandit Interco tensed. I would say that if we have the I m own positive and I am that's one is smaller than the I am for all m and then doesn't implies that the submission off that I am it will have the same behavior on the integral. On that, I asked the X This will be the subscript X here and now in this question, we even the submission off the two off three n plus five from one to infinity here to applied in protest. First of all, we need to compute in the grow from one to infinity. And now we have the 2/3 now and we tend to be the express five, the X and this in the grow, it will equal Jew. The Jew terms the island off the three X plus five and by the general. We need to divide it by three here and evaluated under one to infinity. Therefore, we should get equal Thio Thio on top three And now we have the airline and infinity, ico to infinity and the island off and then minus and land off the one in Segunda eight and still is an echo to infinity. Therefore we can conclude that list is here will be divergent. Go

The series converges or diverges, I'm gonna Okay, here's the submission of Helen of and overhead And starts from 2 to infinity here. So first we're going to confirm whether or not we can use the to protest. So first off Helena bonham brand friend going to to to infinity is definitely great in zero, so it's positive. So that checks out next is if it's continuous, which it definitely is Ellen Evan and and here, since it's going to be two onward and everything that can be equal to zero. So therefore it is continuous And then the last part is if it's decreasing, so we have found enough to over two than Ellen of 3/3, right, and we check those out.

Kids trying to check and see if we gave the integral test on this 1st. Let's get this information form This for an equals one to infinity of Hell enough. And plus one. Well friend Plus one. Okay, So I can definitely see that this is positive and continuous because the what would make that a number equal to zero would be an equals negative one. So that's true for end screen or equal to one here, spoken on them and then the next part. Let's check and see if it's decreasing. So let's go ahead and take the trade of this. So driving this here we've got she have prime minus yes cheap prime here all over she squared so we got negative hell enough plus one. All right. This one squared bottles always positive. Therefore top is definitely less than zero Friend Creator Ankle to one. So it is decreasing. Okay. All right. So from here, what we're gonna do is going to use the integral test here. Do one from to infinity. Well enough and plus one Of her and plus one here. And if we use U substitution for alan and plus one we get to you is equal to one over plus one here. So therefore we have the integral of you chance of being U squared over two. So something that in you get LNFN plus one squared and taking that from one to infinity. And when we apply affinity to that, it goes to infinities, therefore the series to the richest.

Trying to confirm that this hmm that we can use the integral test for this. First thing checking force we have Helen of an over radical and from radicals want affinity here. So definitely positive. See that So it is positive here For all values event, greater or equal to one. However, when we look at the so definitely is continuous here. About that we've got a natural log and the square root function, both of which are continuous. So that's fine. Final part which is whether or not it's decreasing. Look at the value of Helen of two over radical two. It's actually So that's .49 approximately. Helen of 3-hour radical three Is .63. A lot of four radical for it 62.693. So we see that it's actually increasing so cannot yes integral test as a result of that.


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