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Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?$ a_n = \frac{1 - n}{2 +n} $
Bounded and monotonic
Calculus 2 / BC
Chapter 11
Infinite Sequences and Series
Section 1
Sequences
Series
Oregon State University
Harvey Mudd College
Baylor University
Idaho State University
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let's determine whether this sequence is increasing or decreasing. Or perhaps it's neither. In which case we say it's not monotone. Then we'LL see whether or not it's founded, so to show that it's monotone well, we can do is try to show. For example, if we'LL claim that it's decreasing, which will be the case here, then this means that that the following term is no larger than the previous her for all in. So we can go ahead and show this by algebraic Lee by hand, or we can look at the function. Sometimes this is just more convenient to use. Let's go ahead and use X so we can use some calculus, take derivatives and so on. Then here we're defining f such that F event equals an so to show that the end sequence is decreasing. We'LL just show Epps decreasing and to show ifs decreasing well to show that the derivative is negative. Okay, so let's go ahead and compute that derivative f prime of X. We're using the quotient rule here, So you recall the question rule don't. So we go ahead and take the derivative of the numerator so it's minus one and then times the denominator minus the numerator times, the derivative of the denominator over the denominator squared. Now that numerator loose that's minus two, minus X minus one plus x So those exes will cancel and I have negative three over a square, and this is always negative, that denominators positive. But the numerator is negative, so the fraction is always negative. So this allows us to conclude that the sequence and is increasing. So now the next part is whether or not this thing is actually bounded. Let's go to the next page. Since a M is decreasing, we have at the limit of, and if it exists, it's less than or equals who a N is less than or equal to a one. So this is due to the fact on ly that it's decreasing a ones the largest. You're only getting smaller after a one. This is why I am is less than or equal, say one. And since the sequence is continuing to decrease, the limit, if it exists, will be the smallest of them all. And for this problem we have a one that's equal to zero. We can go ahead and find a limited a and just buy using some algebra or Lopez House rule. In either case, this is negative one so that we've just shown that A N is between negative one and zero. This is for all in. Therefore, the sequences also abounded. So going on to the next page to summarize and is bounded and is decreasing, and that's our final answer.
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