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# Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?$a_n = 3 - 2ne^{-n}$

## the sequence is bounded

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##### Catherine R.

Missouri State University

##### Samuel H.

University of Nottingham

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first, let's show that this sequence is increasing, so the way to do that here is we can define f of X to be three minus two X e to the minus X. Now f is monotone or I should say not F, but an This sequence is monotone if and only if the derivative has the same doesn't change sign, let me work it that way doesn't change signs. So either always negative or always positive, but not both. So here, let's look at F. Take the derivative. So here three goes to zero, then we have minus to either the negative X and then plus two x either the minus X. I use the chain rule there canceled out some negatives, so this is either the minus X, and then we have two plus two x here also because we're dealing with the sequence and it's bigger than or equal to one. So that means we want X to be bigger than or equal to one. That's important because here, if X is bigger than or equal to one, either the minus X is always positive for any X. But then two plus two X is also positive due to this condition up here. So we're multiplying two positive numbers together, so the result is also positive. So this shows that a N is increasing because positive derivative means increasing F increasing f means AM is increasing. So that answers the first part of this question. Now we'll go on to the next part to determine whether or not this thing is bounded. So to do that, I'll need more room. Let me go on to the next page, so limit. So here the question is whether it's bounded the sequence. So the limit as n goes to infinity of the end of term. Mhm. Yes. Now the first part. Well, that's just three. And then let me go ahead and push this inside and let me write that limit as X instead of in. So I could use local tells rule if I have to, and we can see here that if you rewrite the e and without the minus sign, you have infinity over infinity and the limit, so we should use low petals role here. So this is by low Patel of that we have a limit of the type infinity over infinity. That's an indeterminate form so low, Patel's rule says. You take the derivative of the top and the bottom, so that's a tool of top. The denominator ends up staying the same. Absolutely differentiate. And then we have two over infinity. So this is zero. So we just have three minus zero equals three, and we could conclude that the sequence is bounded. So we have and is bounded. And the reason why, since its emergence and convergent sequences are always bounded something. So to summarize, we've shown that a and is bounded and monotone increasing mm, mhm, and that's our final answer.

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