## Proving a sequence converges using the formal definition | Series | AP Calculus BC | Khan Academy

Posted in Home Furnishings, Local journalism, Uncategorized on October 24th, 2020 by Mason Fletcher

In the previous video I territory claim that this streak here it can be determined by explicit assigning in a way that the frontiers of the row I can write it as minus 1 to the power of n plus 1 on n. This is one path to characterize a number explicitly the frontiers of this, when n tends to infinity, is similar to 0. Like she looks like this. n is getting bigger and bigger and bigger, although the numerator fluctuates between minus 1 and 1. It seemed to be coming smaller and smaller and smaller. But I haven’t proven it like that i want to do now in the video. To prove it, it would be true if exclusively if for each epsilon greater than 0, has a principal M that is greater than 0, Where if a little n, if the index were higher than the main M, then the nth member of the succession is in accordance with the epsilon range for the limit, within epsilon of 0. What does this convey? This means that our limit is 0. Let me write it in a brand-new coloring. Our limit here is 0. This is the limit. The restraint here is we say that the sequence is convergent to 0. We say that if we are given an epsilon around 0. Let’s say this here is 0 plus epsilon. This is 0 plus epsilon. The behavior we described it now, it looks like epsilon will be 0.5. This will be 0 minus epsilon. Let me sucked it a little clearer. This is likely to be 0 minus epsilon. We have negative epsilon, 0 minus epsilon, 0 plus epsilon. The limit in this case or our statement about their own borders is 0. This is to say that for each epsilon, we must find such M, where if n is greater than M, the interval between the sequence and the boundary will be less than epsilon. If the distance between the sequence and the border is lower than epsilon, that is, the cost of the string for a returned n will be within these two restriction. It must be in that range now, which I colors over a definite n. Therefore, if I select n to be here, it seems to be anything bigger than that will be the case in which we will be within these limits. But how do we prove it? Let’s just think about what it has to happen for that to be true. What must be true to be a with indicator n minus 0, the absolute evaluate of a with indicator n minus 0, to be less than epsilon? This is another way of saying that the absolute value of a with indicator n must be less than epsilon. and with indicator n is right that now, so this is another way of saying the absolute appreciate minus 1 to the power of n plus 1 over n must be smaller than epsilon, which is another way of saying because this is minus 1 to the power of n plus 1, this numerator is simply replaced by a negative one and a positive edition of 1 over n. But if you calculate the ultimate ethic of that, it will always be positive. This is the same as 1 over n, as the absolute importance of 1 over n must be less than epsilon. n will always be positive. n starts at 1 and goes to infinity. This evaluate will always be positive. This signifies the same as 1 over n it has to be less than epsilon for this thing to happen here to be true. Now we write the reciprocal on both sides. If we make the reciprocal on both sides for an inequality, we will have n if you make the reciprocal on both sides of an inequality, you exchange his ratify. For this to be true, n must be greater than 1 on epsilon. In essence, we have now proved it. We said that for this particular number, if you give me some epsilon, I will i knows where to find M which is 1 on epsilon. Because if n is greater than M, which is 1 on epsilon, we will know that this will be true here. That will be true. So their own borders obviously exists. So here, for that definitely epsilon it seems that we have chosen 0.5 or 1/2 for our epsilon. As long as n is greater than 1 over 1/2, which is 2, so in this case we can say that if you gives people 1/2, M will be a function of epsilon. This will be determined for each epsilon that is larger than 0. 1 on 1/2 is over there. I’ll prepare M be here. You learn, this is really the case, in which the array is within the range, when passing through any n larger than 2. For n equivalent to 3 it is in the series. For n equivalent to 4 it is in the straddle. For n equal to 5 as it continues. We proved it now. We did the proof. If you give me any other epsilon, I said M is equal to 1 on this thing. For any n greater than this, this will be true. Therefore, this is definitely the speciman. This sequence is similar to 0..