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A n = 1 E 1 sequence ifit converges; otherwise indicate divergenceDiverges...

Question

A n = 1 E 1 sequence ifit converges; otherwise indicate divergenceDiverges

a n = 1 E 1 sequence ifit converges; otherwise indicate divergence Diverges



Answers

Determine whether the sequence converges or diverges. $$a_{n}=\frac{e^{n}+2}{e^{2 n}-1}$$

So for this problem we have a sequence described by the equation is of em is equal to erase the one over end and what we wanted to identify and this problem is to determine whether the sequence will converge or not and if it does converge, we wanted to identify what is the limit. So let's do a quick check for something values for a few values of N and let's identify the hardest funding is of and for these end values to for example, when N is equal to one will have is one which is equal to erase to 1/1 Or simply erased one And that should be equal to 2.70. Okay, let me just quickly check that 2.72 When N is equal to two, we have erased one hand And it is one hap Is equal to 1.64 Roughly 1.65 When N is equal to three we have Here is 1 3rd or approximately 1.40. And then lastly when N is equal to 14 to 4, A seven is equal to It is the 1 4th Which is approximately equal to 1.20. Yeah, and as you can see here actually try to have a larger value of and the value of this event is getting smaller and smaller but we are not sure what is the limit of this valley of S F N will be. So for us to analyze this let's let's quickly look at this portion of the case of an equation they killed that one over in Is getting smaller and smaller as N approaches to zero. So as an approach as N approaches infinity. Once one over end approaches zero and we can actually confirm that went in is equal to one. One over in is just 1/1 or point When it is supposed to this is 1/2 or 0.5 When N is equal to three, this is 1/3 or .33 and it will try to plug in a higher value of ends a 10. This is 1/10 or equal the point point and if we try to increase that some more We'll have one over 100 here or .01. And as you can see us in gets larger and larger. The value of one over and becomes smaller and smaller, approaching a value of zero. So that being said, if I wanted to find out what this event is as N approaches infinity, then it is simply equal to e race zero because I know that one over end approaches zero. As N approaches infinity And take note that any number faces zero is always equal to one except accept it. When the base is zero or infinity. So any number except zero and infinity when raised to the power of zero, it will always end up having a value of one. So for this problem, we could say that this sequence will converge to a value one, It converges to a value of one

In this exercise, we come to the side if this statement is true. So this tainment looks very similar to the comparison Direct comparison test. But it's a slightly different. So he's saying that if a n it's a smaller regaled and being and be in d Burgess diver, choose then also a in my burgers. So in the case off the direct comparison tests, where we have is that bnb? Yeah, converges then a convergence. This is slightly different. And the truth is that this statement isn't five in fact, false. Okay, let's see an example. For example, if you consider a and to be one of Ren Square one over and square is the convergence sequenced in and be over in to be won over an isa divergent series. Now siro is always bigger required on one over end square that is so is smaller equal than one program. So this is a counter example because we know that the serious being diver just and siris a n converges

Once again we're looking at limit as n goes to infinity of Anne. If this limit exists and is finite in the sequence converges otherwise it's said to diverge. Since the exponential function is continuous, we're allowed to write this as e to the limit as n goes to infinity of minus one over squared of end. You're the continuous function. You can pull the limit inside of the function. So that's what we're doing here. And now this. This is something that we should know how to evaluate his in, goes to infinity, squared of and is going to go to infinity. So we're going to be looking at minus one over infinity, which is just like zero. So this turns into E to the zero r E to the minus zero, and that's just one. So this converges two, one

In this question were given that the summation off some and and this one from one to infinity, it will be convergent. And by the're, um, doesn't implies that we have the limit on the high end Angus infinity every equal to zero and therefore we can say that they exist. Some capital in here says that for on am greater ICO to you and and then the I am here will be smaller than one right? And then when any small and smaller than one we consider the p n Echo to the one over I am and for on and greater equal to capital and hear this end and listen, it will be, ah for sure. It's greater than one because in the end, smaller than one So one of us something smaller than when we bigger than one. Therefore, we can get the limit off the p n. It's industry. Infinity, it will be, is not equal to the zero and doesn't implies that we have the submission on the PM from one to infinity. Excellent coaching assumption of one of I m. From one to infinity, it will be divergent


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