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Lct the sequence (Tn be defined recursively by%1 12 Tn = 1+ In_1 ifn > 2Prove that (Tn) converges, and evaluate the limit....

Question

Lct the sequence (Tn be defined recursively by%1 12 Tn = 1+ In_1 ifn > 2Prove that (Tn) converges, and evaluate the limit.

Lct the sequence (Tn be defined recursively by %1 12 Tn = 1+ In_1 ifn > 2 Prove that (Tn) converges, and evaluate the limit.



Answers

Show that each sequence converges. Find its limit. $$ \left\{\ln \frac{n^{2}+2}{2 n^{2}+3}\right\} $$

In this question. We need to determine whether the sequence and Square -1 upon two and converges and we need to find its limit. Let's see how to solve this question. Consider a and if he calls to Any Square -1 upon to win, in order to find the limit of the sequence, we can write limit and tends to infinity A&F equals two limit and tends to infinity and square Pakistan upon to win. We can write it as limit and tends to infinity and esquire into 1 -1 upon any Squire upon to win. The one end will be cancelled with one end hands. This will be close to limit and tends to infinity and by two and 2, 1 -1 upon an Esquire. And this can be routine is limit and tends to infinity and by two into limit and tends to infinity 1 -1 upon any Squire. No substitute and every day on the place of n. So this will be calls to infinity upon to into 1 -1 upon Infinity Squire. So this will be called to infinity the airport, we can conclude that the sequence and the Square -1 upon two and diverted juice and yeah, the limit is infinity. This is the final answer for this problem. I hope you understand the solution. Thank you

Hello. So here are given sequence is the quantity to m plus one squared over the quantity three and minus one squared. So the check whether this sequence is going to be convergent. We take the limit as N tends to infinity. Well we can go ahead and square both the numerator and denominator out. So then we get the limit as an tends to infinity of a four and square plus four and plus one and then divided by a nine N squared minus six N Plus one. Okay. And we can go ahead and divide the numerator and the denominator um by the highest powers, divide everything by and squared. And then we end up here with the limit as N tends to infinity of just a four plus four over N plus a one over and squared and then all divided by nine minus six over N plus a one over and square songs. And it goes to infinity that goes to zero, that goes to zero, that goes to zero, that goes to zero. We're just left with 4/9. So therefore, yes, this sequence does converge. We are convergent and we converge to the value of 4/9. Mhm. Take care.

Compute the limit off the land and the studio over and square ministry. It's an interesting infinity in at this time we can push the limit inside the island. Therefore Ghenda and land off the limit. Angus Infinity on the interest. You are over and square ministry and you can be tediously emit here. I will rebuild the Ellen outside. I will kept me inside. Now I will factor that and square outside. I didn't have the one off and press chill over and on. The bottom had the end square. And then we have one minus three of and square and I will say we can Cancer the and square with and square. It's an industry infinity doesn't just use. There are generous juicer as well. And isn't this Juzo? Therefore we have left with the end, and I'm no is over one and a cogent. And I'm, uh uh disagree cogent and re zero and go to the minus Infinity here. So doesn't be divergent.

Hello. So here we have our sequence. Um which is given as two times the natural log of three n minus. Natural log of n squared plus one. Okay, so, well, the check for convergent or divergent, we take a limit as n goes to infinity. Um, so what we get then is the limit as N tends to infinity. Okay. Of this sequence gives us um, just a nine squad. The natural log, mm hmm. Um right. We can write this. We have a difference of two lobby. Right? This as the quotient of a single log. And we get the natural log of nine and squared over and squared plus one. Okay? Um, then, well, we can divide everything by our highest power. So then we get the limit as N goes to infinity of the natural log of well divided by n squared. We get a nine over a one plus one over and squared. And now it's N goes to infinity. One over and squared. That goes to zero. So therefore it's gonna be just the natural log of nine. So yes, our sequence is going to converge and it's going to converge to the natural log of nine. Yeah.


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