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Use ne convergent: 8 n2 integral Test to determine whether the following seriesThe The The None Test Test the function of the does does Test apply apply options doe...

Question

Use ne convergent: 8 n2 integral Test to determine whether the following seriesThe The The None Test Test the function of the does does Test apply apply options does and but not decreasing: the the not series apply improper because convergent: integral the first diverges: derivative of flx) is positive

Use ne convergent: 8 n2 integral Test to determine whether the following series The The The None Test Test the function of the does does Test apply apply options does and but not decreasing: the the not series apply improper because convergent: integral the first diverges: derivative of flx) is positive



Answers

Use the integral test to determine whether the series is convergent or divergent. $$\sum_{n=1}^{\infty} n e^{-n}$$

We want to evaluate the convergence of the given series. Using the integral test. The series in question is a some from an equal studio infinity of one over N times the square root of L N N. So this question is testing our ability to apply convergence tests for series. Specifically the integral test. Let's define what the integral test is before we proceed to solve. So V integral test states that if the definite integral from K to infinity fx dx converges so too must be some from N equals K to infinity of AM we will write A M as F X. Dx. So it we also have to have it at that before integral diverges. So too much the series. So let's first rewrite AM as F X D X and then solve the integral. So we can rewrite A N one over N route L N N as D X over X. Root L N X. This integral has anti directive to root L N X, evaluated from two to infinity. We have two element ruutel infinity minus two L into the first term shoots off to infinity. So the integral diverges, which means that our series also dive.

We want to use the interval test to determine whether this Siri's converters or diverges. So they tell us to check the conditions to ensure that internal testes satisfied. So let's first go ahead and check to see if our function is continuous. Our that's positive is the first thing we should actually check. So let's go ahead and call. No, our sequence there. So natural log of in squared over in. Well, first, we could rewrite this us two times the natural log of in over in. And I'm just gonna want to rewrite it like this for when we check to see if it's decreasing car, not some when we take the derivative of it. Well, if we're plugging in values larger than two natural log into is positive in is also going to be positive. Divide to positive numbers. So this is going to be strictly greater than zero. Now we need to check to see if the function is continuous and the only issues will have is when we divide by zero. So since and his large and the two no issues there and for negative values, because remember, the natural law could only plug in numbers larger than zero and since and is larger than zero, it's also good. So to continues, functions divided by each other at least on the interval, larger than two works out. And now we need to check to see if this is decreasing. So we will go ahead and find the derivative of this so f prime. So I'm gonna take the derivative of this one over here so we can use quotient rule. To do that, there's gonna be two times so low d high. So one over, n minus high de lo all over the square. What's below? And now, if we were to go ahead and simplify this down would have two times one meister natural log of and all over and squared. And remember, this is for in greater than or equal toe to so we can make sure that this will be negative if our numerator it's negative. So this is lessons because in squares, always positive. So we would add natural log cabin over. So we have one greater than natural log of in, and then we exponentially it inside. So we want to make sure that in is larger than E Qari and so he is about 2.71 So that means if we take in to be three, then this will work so we can go ahead and just let nd three. And in this case, well, we can come over here and rewrite our Siri's to be. So we first plug in to so I'd be natural log of or over two and then some from and is equal to three to infinity of natural log of in I'll just due to natural event over in. And now this is still satisfying it being positive and continuous. Because on this interval, here is where it's positive and as long as we have in three here, then we have that it's decreasing on this whole interval. So this here will check out. So now we can go ahead and apply the integral test toe. This Siri's here and so is going to be three to infinity of. I'm gonna factor that two out so two times the natural law given over and now we can integrate this using a u sub and it looks like the substitution we're gonna want to make is gonna be you. Is he going to the natural log of it. And so then D you is going to be won over end Deion. So then that means we can go ahead. Every right are integral here as two times the integral oh so natural lago three and then the limit, as in a purchase. Affinity. Ah, natural log of in that should go to infinity still So our upper bound is still gonna be infinity and then we just have you, d you know, interpreting that we need to power rule So we'd have you squared divided by two. So that two out there who can't sell so we'd have you squared evaluated from the natural log of three Have you squared, Evaluate from natural of the three to infinity. And so if we were to go ahead and write This is the limit though, So the limit as be approaches infinity. So remember replacing infinity here with B So the B B squared minus the natural log of three squared Well, b squared diverges to infinity Natural log of two squared Our national August 3 squared is just a constant so that is a matter So this diverges and so that implies that are Siri's and is he going to to infinity of the natural log of in squared over in diverges as well?

We want to evaluate the convergence of the following infinite series. Using the integral test, the series in question is the sum from n equals one to infinity of N times E. Raised to the power of negative and square. This question is testing our ability to test the convergence of an infinite series specifically using the given test. So before we saw let's define that test. So we understand the problem. Remember that the internal test states that if the integral from K to infinity fx dx converges. So too does the sum from N equals K to infinity of 80 yen. Conversely, if that definite integral diverges so too must the sound. So we transform A. M. Into the forum F. Of X. Dx and evaluate the integral. So we have the integral from one to infinity of X to the negative X squared dx. Taking the anti devoted. This is just one negative one half of the negative X squared from one to infinity. So this is negative one half of the negative infinity minus negative one E. The negative affinity is just one overeating infinity or zero, so this is negative one half zero minus one over E. Or one over to E. Since our integral diversion, or rather since our integral converges to one over to eat. So too must our series, our series converges.

We want to use the interval test to determine whether the Siri's comfort is or divergence. And they say we should probably also check that the conditions to actually apply, then a protest apply. Let's go ahead and do that off on the side here, so I'll go ahead and say that F is equal to E to the negative to it. So this is strictly greater than zero on all intervals of real numbers. So it doesn't even matter for that animal that we have over here. So that checks out. And we also know this function is continuous for all values as well. Now, to figure out if it's decreasing, we could take the derivative that's gonna be negative to e to the negative to win. And, well, if e to the negative to win was positive for all values that negative two times eats a negative to and should be negative for all values. So we have. It's strictly greater than zero continuous and decreasing. So this checks out. So now we go ahead and integrate. So I'm gonna write this as the limit, as be approaches infinity of the troll of one to be of e to the negative to Ben, do you? And now integrating heated negative to in. That should give us so we divide by its power. So that would be negative. 1/2 e to the negative to it? No, we evaluate from one to be so I'm gonna factor that negative 1/2 all the way helps. And then we're plug and be so negative. Two b minus e to the negative too. Now we know as B goes to infinity, this is going to go to zero. So we're gonna be left with 1/2. He too negative too. So since are integral here converges. That implies that's some mation from in Is he would want to infinity of e to the negative to win con urges.


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