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The 3 sequenceincreasing bounded convergent deceasing...

Question

The 3 sequenceincreasing bounded convergent deceasing

The 3 sequence increasing bounded convergent deceasing



Answers

Which of the sequences $\left\{a_{n}\right\}$ converge, and
which diverge? Find the limit of each convergent sequence.
$$
a_{n}=\frac{3^{n}}{n^{3}}
$$

The formula for series 2 to the power of an over in Cuba. I need to see if it diverges or converges. So let's take the limit as n goes to infinity of too sort of power van over and cute. So numerous is going to go to infinity but so is the denominator. There is an indeterminant form. So since it's indeterminate, we're gonna have to use l hospitals rule which means we have to take the derivative of the numerator and denominator and that's going to leave us with limits as an goes to infinity. A derivative of numerator which is just two to the power of and times Atlanta too All over three and squirt. But again they're going to get infinity over infinity. So we have to repeat hospitals again meaning take the derivative of both top and bottom one more time. Certainly. What's with to to the end times Ln of two squared over six in but this is still not enough. You still have infinity over infinity. So we've got to do this for one last time and that's going to leave us with two to the end Times Ln of two. You all over six. Now this time you're going to get infinity talk here And just a constant six down here. So definitely was fit and infinity which means our series is going to diverge to positive infinity

And is equal to three e to the end, minus E to the minus and over into the end, plus three d minus. And that's the sequence. Convergent diverge. Well, case. So we want limit as n goes to infinity. Aye, to the end. So first, let's observe that e to the minus end is the reciprocal off to the end. And as n goes to infinity e to the end goes to affinity, right? So we can rewrite this as limit as ex ghost or affinity. So here we let x be equal to it to the end. So this is the equivalent of, say, limit as ex closer affinity three x minus one over X over Thanks plus three over X right. And then you divide your numerator and denominator both by X. This is equal to limit as exposed to infinity three minus one over X square over one plus three over and square. Now, as excuse affinity, this one over X choirs from zero and this three over X were also goes to zero. So the result is secretive. Three. The sequence convergence

In discussion. We are required to conjecture the limits of the sequence Under all three. Under road three. Rules three on the road to the Rule three, Rule three and so on. Next question the first time even It recalls 203. Second term eight with calls to underwrote three. Route three. The third term A three equals two. Under Route three Road 3 Road three. Now To establish the record in formula we can write a two equals 2 road three. And for this Rule three which is inside the underwrote we can substitute even So this will be cost 203. Even A three will be called to understood three road three, Rule three. since Rule three, Rule 3 vehicles to A two. So it will be close to understood three. A Two hands on the basis of the sequence, we can conclude that the recording formula A. N. Plus one will be close to underwrote three A. N. Now consider that Mhm. The limit of the sequence is L. Therefore we can write limit and tends to infinity A. N. Equals two L. And this is he cure into limit and tends to infinity e. n plus one if he calls to L. Since we have already arrived that A and plus one equals two On the road three a.m. Therefore we can right limit and tends to infinity A. N plus one is calls to limit and tends to infinity on the road three in. So the silver calls to limit and tends to infinity A and plus one, it recalls to under road limit and tends to infinity three A. N. So the Sylvie calls to underwrote three limit and tends to infinity. And since LTD and tends to infinity and this one is called to L. And limit and tends to infinity and it's called to L. Therefore we can right L equals two under road real now square the both sides do we get and it's politicals too real or the cell vehicles to I require minus Really because 20 hands Early in two L -3 because 20 sins. The first term of the sequence is under three and all the terms are positive hands. The sequence converges to on the road two. Therefore they limit of the sequence L equals two limit and tends to infinity A. N. Equals two Under 2. So this is the final answer for this problem. I hope you understand the solution. Thank you.

The sequence we have here is three plus. Eat the power of negative to end. You were asked if this is convergent or divergent. So let's take a look at what happens when under approaches infinity. So the limit as and approaches infinity for the sequence is going to be equal to. So the first thing is just a. Three. So we can take that out. So this is really three plus the limit as under purchase vicinity of E to the negative 21 but we can see that this is going towards zero. So which means this is three plus zero which is equal to three. So therefore this is convergent and it converges just three. Mhm.


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