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Let An be the interval (0, 1 /n). Provide a formal proof that n=I An is empty...

Question

Let An be the interval (0, 1 /n). Provide a formal proof that n=I An is empty

Let An be the interval (0, 1 /n). Provide a formal proof that n=I An is empty



Answers

Prove that $a_{n} \rightarrow 0$ iff $\left|a_{n}\right| \rightarrow 0$.

Okay and this question We are given that a north divided by n. Plus one. Plus a one divided by N. Plus and so on. A. N. Is equal to zero. Okay? And we need to prove that the function A north extra depart en plus even extra depart and -1 and so on. Plus A. N. Is equal to zero. Dysfunction has at least one real root between zero and one. Okay so let's take this to be F dash X. Okay because we have according to engagement value tomorrow we have at least one real root of F dash X. Okay so we have taken this to be uh F F dash X. Now fx becomes they're not divided by n plus one extra departure and plus one less anyone divided by an extra the far end and so on A. N. X. Plus some constraint be ok is equal to zero. Now if you check the value of F. Of zero. Okay if we check the value of F. Of zero it comes out to be deep. Okay. And if you check the value of F one, okay, I for one comes out to be a not divided by N plus one Plus A one divided by N. And so on A m Plus d equals to zero. Sorry? Plus D. Okay, this is the value of F of one. Now, if you see this town is given to the zero, Okay. This time has already given the question to be equal to zero. So a +41 comes out to be zero plus D, which is D. Okay, if you know this F of zero F of zero Comes out to be equal to F1, which is equals to be so by roles here. Um By rules tour um we can clearly see that F dash X has at least at least one real root. Mhm Between zero and one. Okay. Because F of zero is coming equals two B f of one. So F dash X will have at least one real root between zero and one. Thank you.

For this problem, we have given the following statement that says is seven is approaching the zero and there is another sequence piece of end that is bounded. And we wanted to show that a seven times myself and will also approach zero. So what I could do here is that if um if this event let's say the sequence peace event is founded, then we can actually say that the limit of visa end As an approaches zero is some value X. And its limit the limit of the sequences and approaches to infinity is some value. Why let's say because it's bounded, it's bounded between X from X to y. Now, what I wanted to do here is to evaluate next, this is the first thing I did evaluate E N times piece up in as SFN is approaching zero. So how how will I do that? I'll just get limit of S F N B C f N as N approaches to zero. And um we know hold on As S. Sorry, I'm evaluating it as SFN is approaching 2, 0. Okay, so we have two cases here. So for case one, if a seven approach to zero, when In is approaching the zero, then the limit of piece of in as an approach to zero, the limit of a seven times visa then is equal to zero times X. Because we recall that as N approaches zero, the value of peace of end sequel to this. Okay. And when a is out and this is the first case, we say when this event is approaching zero. Uh that happens when N is approaching 20. So this is our case number one, Which is equal to zero. So the other case key somewhere to if is of and approaches to zero when and is approaching infinity. And we know that when N is approaching infinity, the value of peace of end is approaching two. Why then cops and aren't, then we can say that the limit of S F N times P seven as A seven approaches to zero equals zero times. Why? Which is the value of peace of NsN approaches to infinity And that is still equal to zero. So that being said in either case is NBC event is approaching zero and we were able to Yeah.

Okay. So we can start by pulling out our power. So we gets and log base A of a among us. May of a is equal to one. So this is it going to end?


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