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Let (an)nzl be a sequence of real numbers and suppose that an aeR Let f :AvRbe & function. For each n e N define Jn:A-Rby Jn() = fW)+an: Prove that conve erges ...

Question

Let (an)nzl be a sequence of real numbers and suppose that an aeR Let f :AvRbe & function. For each n e N define Jn:A-Rby Jn() = fW)+an: Prove that conve erges uniformly to the function f +aon the set A (For example, fn(x) = e+4 converges uniformly to e on R.)

Let (an)nzl be a sequence of real numbers and suppose that an aeR Let f :AvRbe & function. For each n e N define Jn:A-Rby Jn() = fW)+an: Prove that conve erges uniformly to the function f +aon the set A (For example, fn(x) = e+4 converges uniformly to e on R.)



Answers

Let $\left\{b_{n}\right\}$ be a sequence and let $a_{n}=b_{n}-b_{n-1} .$ Show that $\sum_{n=1}^{\infty} a_{n}$ converges if and only if $\lim _{n \rightarrow \infty} b_{n}$ exists.

Yeah. So in this problem we need to sure that any bounded decreasing sequence of real numbers converges. So we are given about a decreasing sequence. We named that A. N. And we need to show that A. N. Can gorgeous to what limit? We don't care about this. Okay, so let's start with the proof, send the sequence in is bounded in particular. It does bounded below that is the set A. And such that and comes from natural numbers is a bounded subset five. Have we known that are has completeness property also known as and you'll be property that is least upper bound property. And this is equal into G. L. B. Property also known as greatest lower bound property, basically this means, but if any subset of R is bounded below, then it must have an infamy. Um So in fema of the scent and coming from and coming from national numbers is some real number. Let's call this L. So our claim is like the sequence N. Can we just to L now, since L is the greatest No, it bound after said of elements of sequences of the terms of sequences. So for any absolutely. Didn't see role as minus epsilon, which is strictly less than N. I'm sorry, L plus epsilon, which is strictly created. An L cannot be by lower bound. So there exists some. And in particular there exists some term A N. That is less than L plus of silence. But since in is a decreasing sequence we have a N less than equal to capital and for every and greater than or equal to capital. And so in particular we have and less than L plus a silent for every angry to an end no sense. And is can feel them. We have less than equal to a N. For every n natural numbers. In particular, l minus epsilon, which is strictly less than L, would also be less than a N for every angrier than capital. And so we have this inequality right here and this inequality right here and we combined them to find that for every epsilon created in zero, there exists a national number and such side for every end. After that comes after that natural number added minus upside and is less than a. In which is less than L. Plus of silent. That means A. In the sequence A in Canada. Just to L. This completes option for this term.

A role for this problem is to determine where the function is continuous or where it isn't continuous. So um it's important to know that in order for a function to be continuous, the limit as X approaches C of F of X is equal to F F C. So what that means is that if we have something like X squared, we know this function is continuous at X equals zero because the limit as X approaches C is zero and that's the same value as the function at sea. So we know that that is where the function will be continuous. However, if we had something like this right here, then what we end up seeing is that this function is not continuous because the limit as X approaches here is going to be one of the limit as X approaches here is going to be negative one. So the limit doesn't exist and we see the function has these different values. So that's how we make sense of continuity.

This problem covers limits. In order to solve part A and B of this problem, we will use the approach of contradiction. Let's look at part A. First. The claim is that if S. N. Is greater than equal to A. For all but finite lee many N than limit, S N is greater than equal to A. Let us assume for a contradiction that this limit Yeah, is less than let us assume. And let's say absalon is equal to a minus S. This is the assumption. By the definition of limits, we can say that. Bye. Definition of limit. We can see that there exists capital N. Substantive small and is greater than capital in then sn minus S is less than epsilon. This is what the definition says. I'll write it down there exists capital and such that Yeah. If and is greater than capital N. Then and send minus S. Yes, less than absolute. So if you open this inequality, you know, you will get minus epsilon Nice between s and minus S taking minus s on both sides. You get s minus absalon as less absalon. Now, you can see here that S plus epsilon is nothing but A. So you will use that factor and I can say that s minus epsilon. Uh it means that essen belongs to s minus absalon. Call my A. What does this mean? It means that S. N. Is less than a, S. N is less than A. But our problems said that our assumption was based that S and will be greater than equal to way, but we got the opposite. It means that for all but finite lee many end S. N. Is less than in this problem, as per the assumption, which is a contradiction. It means that our initial claim was correct that if S. N. Is greater than equal to A. For all but financially many in limit S and will be greater than equal to It similarly will solve part B of the problem. Part B says that if S. N is less than equal to be for all, but finitely many end then limit essen is less than equal to be. Let us assume again for contradiction. We will assume, Oh, that limit essen is greater than B. We're assuming as a contradiction. And let us take absalon as S minus B. Again, by the definition of limit there exists and such that if N is greater than in the sm minus S is less than absolute, just like this. We use their definition of LTD so we get yes and minus S is less than epsilon. Again, opening the inequality minus epsilon, S n minus S rather than absolute, this becomes s minus epsilon. Essen? S less. Absolutely. From here, we can see that S many cepsa loans nothing but B. So I can see here, B essen less than s plus absalon. So from here it is clearly seen that sns greater than B. Or I can say for all. But finitely many N. We have S. And greater than B, which is a contradiction according to our statement. Because we took us and less than equal to be. Hence the claim that if S. N. Is greater than less than equal to be for all. But finitely many end then limit us and less than equal to be was correct. And contradiction proved it false. Let's go to part C. Part C is a direct conclusion from part A and part B from part A. You know that? Yes. Which is the limit limit of essen is greater than equal to A. And from part B. This limit S it's less than equal to me, so S belongs to it to be. That's all. Yeah.

In this question here were given. Uh, I am equal to the B N minus. B a minus one. And we're interested in the submission that I am from one option infinity. So notice that they will explain this one when any could you want him to be one minus B zero, then plus B two minus, uh, be one plus the B three months, B two plus the big four minus P five plus up to the PM minus P m minus one. And so, um, and we see them, we can console the B one window. Be one here, be to win the B two B 300 b three. So we're missing the beat three years. So sorry. So let me in. Center doesn't before, uh, be 1234 minus b three here. And then we again this one will be canceled with this one and so on. And here we see that we will cancel this and so on on. We see we have left with only, uh if we consider only up to in the past with some s and for now, then we just stop up to here and then we see that this s and we ego Thio actually end here on me and then we see that this sn it will equal to the We have left with the minus p zero then plus with a p n. And we see that this is here as an we turned the limit here Angus t infinity And then he coaches limit on the industry Infinite. They on the minus B plus B n And we say this one we get echoed you This one will be p zero. So it will be the minus p zero and this leap plus the limit on the PM and goes to infinity. I doesn't implies that the I the submission on the I m converge even only if this limit here also. Uh, I already mitt because you some constant else model that infinity


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