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# Dowód na zbieżności ciągu

Transkrypcja filmu video (w języku angielskim)
I made a claim that for this sequence-- and this was in a previous video-- that for this sequence right over here that can be defined explicitly in this way, that the limit of the sequence-- and so I can write this as negative 1 to the n plus 1 over n. That's one way of defining our sequence explicitly-- the limit of this as n approaches infinity is equal to 0. And it seems that way. As n gets larger and larger and larger, even though the numerator oscillates between negative 1 and 1, it seems like it will get smaller and smaller and smaller. But I didn't prove it, and that's what I want to do in this video. In order to prove it, this is going to be true if and only if for any epsilon greater than 0, there is a capital M greater than 0 such that if lowercase n, if our index is greater than capital M, then the nth term in our sequence is going to be within epsilon of our limit, within epsilon of 0. So what does that say? That says, hey, give me-- our limit is 0. Let me do this in a new color. So our limit right over here is 0. That's our limit. So our limit right over here is-- we're saying the sequence is converging to 0. What we're saying is, give us an epsilon around 0. So let's say that this right over here is 0 plus epsilon. That is 0 plus epsilon. The way I've drawn it here looks like epsilon would be 0.5. This would be 0 minus epsilon. Let me make it a little bit neater. So this would be zero minus epsilon. So this is negative epsilon, 0 minus epsilon, 0 plus epsilon. Our limit in this case, or our claim of a limit, is 0. Now, this is saying for any epsilon, we need to find an M such that if n is greater than M, the distance between our sequence and our limit is going to be less than epsilon. So if the distance between our sequence and our limit is less than epsilon, that means that the value of our sequence for a given n is going to be within these two bounds. It's got to be in this range right over here that I'm shading above a certain n. So if I pick an n right over here, it looks like anything larger than that is going to be the case that we're going to be within those bounds. But how do we prove it? Well, let's just think about what needs to happen for this to be true. So what needs to be happen for a sub n minus 0, the absolute value of a sub n minus 0, what needs to be true for this to be less than epsilon? Well, this is another way of saying that the absolute value of a sub n has to be less than epsilon. And a sub n is just this business right here, so it's another way of saying that the absolute value of negative 1 to the n plus 1 over n has to be less than epsilon, which is another way of saying, because this negative 1 to the n plus 1, this numerator just swaps us between a negative and a positive version of 1 over n. But if you take the absolute value of it, this is always just going to be positive. So this is the same thing as 1 over n, as the absolute value of 1 over n has to be less than epsilon. Now, n is always going to be positive. n starts at 1 and goes to infinity. So this value is always going to be positive. So this is saying the same thing that 1 over n has to be less than epsilon in order for this stuff to be true. And now we can take the reciprocal of both sides. And if you take the reciprocal of both sides of an inequality, you would have that n-- if you take the reciprocal of both sides of an inequality, you swap the inequality. So for this to be true, n has to be greater than 1 over epsilon. And we essentially have proven it now. So now we've said, look, for this particular sequence, you give me any epsilon, and I'm going to set M to be 1 over epsilon. Because if n is greater than M, which is 1 over epsilon, then we know that this right over here is going to be true. That is going to be true. So the limit does definitely exist. And so over here, for this particular epsilon, it looks like we've picked 0.5 or 1/2 as our epsilon. So as long as n is greater than 1 over 1/2, which is 2, so in this case we could say, look, you gave me 1/2. My M is going to be a function of epsilon. It's going to be defined for any epsilon you give me greater than 0. So here, 1 over 1/2 is right over here. I'm going to make my M right over here. And you see it is indeed the case that my sequence is within the range as we passed for any n greater than 2. So for n is equal to 3 it's in the range. For n is equal to 4 it's in the range. For n equals 5-- and it keeps going and going. And we're not just taking our word for it. We've proven it right over here. So we've made the proof. You give me any other any epsilon, I said M is equal to 1 over that thing. And so for n greater than that, this is going to be true. So this is definitely the case. This sequence converges to 0.