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ProblemnLet {sn} be SCQCIC€; of positive reals Suppose thatand < 4 Prove that {Saf...

Question

ProblemnLet {sn} be SCQCIC€; of positive reals Suppose thatand < 4 Prove that {Saf

Problemn Let {sn} be SCQCIC€; of positive reals Suppose that and < 4 Prove that {Saf



Answers

Prove that the statement is true for every positive integer $n$. $$n<2^{n}$$

First but N equals one. The statement is one is monitors and to to power one which is to and this statement is true for any coursework. Next we assume that this statement is true for an equal scape then for a Nikos K plus one it becomes okay plus one yes Mourners and two to the power of cape cross. What? Since this? The statement is true for any calls. Kay and Harris K is monitor than to power Okay. And this is smaller than to the power K plus to the power of cake. Since one is smaller than two to power cake here, K is a positive integral and this equals two times to power Okay. Which equals 2 to the power of K platform, which means the statement is true for ecos K plus one. According to mathematical induction, this statement is true for every and equals a positive integer.

Okay, so here we have that S M N approaches L. Now we have at the limit as N approaches infinity of a suburban, well then it's going to be equal to L. Now we have that the limit as N approaches infinity of alpha times a Saban is going to be equal to the limit as an approaches infinity of alpha times the limit as N approaches infinity of a suburban, which is going to be equal to Alpha times L. So therefore we have that um a alpha um times a seven is going to be equal or is going to going to approach um alpha times out right.

We need to prove that. And the busting two to the end of power. We do this, we use induction. Start with the base case of n equals 11 will be less than two to the first power equals two. So we do find that one is left to. So the base case is good. Now, the next part's a little tree. So then it must be less than to the wait. So given for any K value. Okay, it's less than two today. OK, we have to prove that. Okay, close one is less than two to the game one. So in this way, we know firstly, they k it is less than Kate. Most one. So in transition. Okay, it's gonna be less than two k plus one. So keep going. You take that extra exponents out of the two. So the two of the first gets to buy it over to the K, and once again, you know, the K is less than two. Okay, We also know that k over to it's less than K. So therefore, by this rule, okay, divided by two must be less than to decay. So therefore Thetis is valid for okay, plus one. That's our answer.

Okay. This problem requires a bit of thought. It's different. We're dealing with inequalities, right? Saying that for any natural number And remember, all positive. Introduce that it's going to be a lesson to to the end. So this seems obviously true. Of course, we plug a one in for end, we'll get one is less than two to the first. That's true. Uh, we want to solve by induction. So we'll assume that the K cases true that K is less than two to the cave. All right, then we can look at the K plus one scenario. All right, so we'll have K plus one is less than two to the K plus one. Uh, we can rewrite that right side if we'd like as to to the K times two the generally, you know, we might just get stuck here on this is if if we've just been working a bit mindlessly, you might really be trying to get that factor of two, and we're not interested in that. Generally, and inequalities were trying to solve in this strategy, they're saying, Okay, some number end, that's less than why. All right, And then our goal then, is to maybe find a number X. That's a lesson end so that afterwards we can say, Well, excess an n n is us. And why, than X must be less than why in this case, we don't necessarily have to Interesting equations. We do have positive integers, right? So I means K or N whichever variable we use is gonna have to be greater than or equal to one. Right? So 123 all the positive integers We, uh, actually will use that in this question. So, um, let's see what we've got here. P of k. If we'd like, we could be right that also, um, won't necessarily help us very much. But, um, we could it might get us one step closer. Let's take a look Care. So this is the same as if we multiply on both sides by two. We get to to the K I was less than two to the K plus one right or two to the K times to to the one which is the same as to to the K plus one. All right, now coming back down to here, we're kind of stuck at this point. We don't We're trying to find that middle term. Right? So let me leave this up here. What do you use X and Y X is less than it were kind of aiming for this. If you can get both of these parts of equations, then we're kind of golden at that point. Let's use this K is greater than or equal to one. Um, we Yes, there we go. OK, so our middle term right now is is either K right or to K if we wrote it in this this way. Um and so what will help us is using this two K version. All right, so looking at this, this is just the definition for natural numbers. We have assumed that on the question because they told us that, uh, we could add K to both sides if we wanted to. The reason we might do that. So that under left side, we could get to K. Right. Um, and now you can see we have something valuable. We have one plus K that matches here, and we have to k, which also matches right here. All right, so let's write everything else. Everything that we know. We have one plus K, it was less than or equal to two case. Then we have two K is less than two to the K plus one. Okay, so we could say that K plus one must be less than two to the K plus one. Now, this symbol might be throwing me off a little bit. But let's say even if if we prove that this was equal which is possible in this case, alright, If K plus one equals two to the case, it still is less than two to the K plus one. So here we have shown that K plus one is less than two to the K plus one, which is exactly what we were trying to prove, right. That's that's right here were able to prove that.


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