In this video, I show that the sum of the series with terms (ln n) / n diverges to infinity by computing its corresponding improper integral via the Integral Test. First noting that the natural log...
Transcript
Example 4, and this states determine whether the following series converges or diverges. And then we have over here, the summation from n equals 1 to infinity of lawn n over n. So a lot of ends over there. So remember, lon is just log base e. So solution to this, so the function fx equals lon x over x is positive and continuous for x is greater. than 1. So you just recall the graph of LON. It's always going to be positive after x is greater than 1. And I'll be illustrating that soon. And again, yes, it's positive and continuous, because the logarithmic function is continuous as well. But it is not obvious whether or not F is decreasing, so we compute its derivative. So yes, from my earlier video's logarithmic function, yes, it is continuous. You can see those. in the earlier videos, but again, this function not necessarily is increasing, because this is divided by X term there. So what we could do here, if we're looking at this function version of the series, FX equals Lon X over X. First I'm going to do is recall the derivative of the long function, so recall that D over DX of Lon X is equal to 1 over X, and you can see this in my earlier videos as well. Yes, so thus, the derivative of this function of top so thus let's just write it down f x equals to lon x over x so now it's taking the derivative so f prime of x equals to then we're going to take the derivative of lon x and i'll just write this in more detail like this d over d x of long x I'm going to put a bracket, and then I'm going to multiply by this 1 over x, and then plus the opposite of this, just to apply chain rule instead of using the quotient rule, because it's not as straightforward to memorize. So I'm just going to write long x like this, times it by D over Dx of 1 over x. So what I'm doing right here is just the chain rule. Should maybe put it like this. Chain rule, top there. where you basically take a derivative times the other function, plus then that function times the derivative of the other function. So this is going to be equal to 1 over x times 1 over x like that, plus long x, and now the derivative of this 1 over x, that's going to be Let's just do this in more detail. It's easier to take the derivative as flipping this above, so you don't have a denominator, so D over the x of x to power of negative 1. So then let's just keep going. This is 1 over x times 1 of 1 x, it's 1 over x squared, plus long x. Now the derivative of x power of negative 1, you're bringing down the negative 1, recall from the power derivative rules. or derivative of power functions, so you bring down the negative 1, and then you minus a power. This is going to be minus 1 like that, and then times it by x minus a power, x negative 2, which is just 1 over 1 divided by x squared. So that's a negative, so this becomes, let's see, I put this all together, this becomes 1 minus long x, because you use the same common denominator, so make it minus, this is x squared like that. Yeah, so this is our derivative, f. prime of x is this like that and now to see if this is increasing or decreasing I'm just going to recall on just a refresher on inverse functions and this lawn of x or log base e so recall so if we have a function like this y equals e to the x then the inverse the inverse is such Yeah, such that, if we were just to do, you just switch the variables, it's such that, or just move it down here, it's such that if you switch the variables, we're going to have an x equals to e to the y, then we need to just solve for the y. So if we solve for the y, yeah, we get the inverse of this function. So what we get is, and the solving for the y, this is defined as, all right, defined. defined as the inverse. To define, it's going to be, now, to solve that y, y, equals to log base E of x for the inverse of that function there, like that. Now, this equals to, and then this is just conveniently written as lon, LN, like that, lon of x. So log base E is lawn of x, like that. Yeah, so this is what we have. So essentially, if we were to graph it out, if we were to graph out Y, but we'll end up having as number E to the Y, this X increases more than the Y. So it's going to be looked like the, recall how the function looks like, and it's always going to be positive. I mean, the X is always going to be positive, but this one could be negative initially, and that is actually when X is less than 1. So I'll just draw this out, so we have X, Y, like this. Recall that the curve looks like this. So we're going to have something that looks like extends out, and it's going to go like that. That's how the log function moves. So it moves x-y's more than it does upwards in the y direction because of this E to the Y. So E to the Y makes X increase faster. So at this point right here, this is at x equals 1. So this is at 1. And recall that number, it's the inverse of this function. So if x is equal to 1, then this is going to be e to the 0. Recall that this is just e to the 0 equals to 1. And again, it's by definition, E to 0 is equal to 1. I mean, any powered power of 0, as I showed my earlier videos on that. And now, the interesting part here is, at this point right here, if this is E, that is E, so if X is equal to E, what we end up having as a value here, this is going to be, well, is the same thing here, E to the, E to the, what number, what number here, I'll just put a blank. E to the question mark equals to E, that's just going to be well 1. So, E to the 1 is equal to E. So yes, remember, it's just switched. So we're trying to find Y. So if this is E, E equals E to the power of y, which is 1. So in other words, this value is all the way here. It's going to be 1 like that. Let's draw some arrows like this, and so on, all the way across there. and this function is our y equals to lon of x or log base e of x so notice what we have here so lawn of x so we have the x squared is positive we have one minus lon of x so if we want to know yeah we want to know if it's decreasing so whenever we want to check if it's decreasing so that we could see if it's converges or diverges or we could or or the other way around if it's increasing uh so what we could do is we'll just look at this graph it's going to be one minus this lawn of x but then as a lot of x here is greater than 1 over here. So it's only greater than 1 when x is greater than e. So here, put an error like this. This right here. I'll put a, yeah, just put a bracket here. So this is greater than 1 here. Or I'll just write y is greater than 1 when x is greater than e. like that. So that's in this upper region over there. Let's draw more and more lines like that. Yeah, so thus, let's just write this down. So that's the derivative, let's put this here, thus, F of X, I mean, derivative F prime of X is equal to 1 minus Lon of X over X squared. This is going to be less than zero, or decreasing. Yeah, decreasing when, when we have that, let's put it over here, so when 1 minus lawn of x is going to be less than 0, because this is all positive, and in other words, we just move this around, move the lawn of x over there, we get a, yeah, we get a lawn of x, make it positive, and be greater than, or just make it less confused, just keep the same size, just move this over, so lawn of x like that. and it's going to be greater than 1. Again, take the power of E on both sides. Like that, these cancel, just different properties of the log function. So this cancel we're going to have now, X is greater than E to the power of 1, which is E. Exactly, I just show it here. So Y is greater than 1 when X is greater than E. So in other words, greater than 1, 1 minus a number greater than 1, it's going to be negative. So yes, so over there. Yes, so it follows that f is decreasing when x is greater than e, and so we can apply the integral test, and what we can do is take the derivative, I'm going to take the integral from 1 to infinity of long x over x d x like that. Now this equals to the limit as as t approaches infinity, because the improper integral of of the integral from 1 to t of long x over x d x like that yes and now recall to well or a note to solve this integral it's going to put a note right here a note so note that the and note that the derivative, so if we take the derivative like this of d over d x of this is the interesting part, because we have an x over there and there's a lon x, the derivative of lon x is 1 over x. So note that if we take the square of it, like that, so the derivative of the square would we end up having is something looks like this, we're going to have a 2, 2 times lon x, and then by chain rule the inside, it's going to be well over x like that. so we're going to have 2 times long x times x. In other words, we have this part right here, is just that. Yeah, so that's the derivative. So then the anti-derivative is going to be the opposite. So thus, the anti-derivative is move this around over to the other side. Yeah, move this two over to the other side. So the anti-derivative of, yeah, long x over x like that. This can be equal to the derivative, I mean, the anti-derivist, so long x, and move the 2 below it, divided by 2, and this is squared, like that. And, yeah, for a completeness, it's just an indefinite integral, this is going to be plus C, a constant C. So, we can ignore that for the definite integrals. So what we could do now, so that's a note, and I'm going to do is put this in a bubble, just like that. and I'm going to put an arrow here, so we just, let's write this down here, so we're going to get a limit as T approaches infinity of, now we're going to have that solving in there. It's going to be over there, lon x over 2, so lawn of, yeah, lon x like that, squared over 2, and then from 1 to t. and now let's just put this all the way down here and solve it. Thus, we have, let's put an equal sign, so equals to extend this calculation. So we're going to have, now, limit, as T approaches infinity, and then just plug this inside, so long t over two, squared like that, bracket minus, and then this is one, like that, so lawn of one. Lon 1 squared over 2, like that. So this one is Lon 1. Let's just look at our chart right here. So, Lon of 1, so when x is 1, we have E to the 0. So this is the 0. So this is the 0, so that's the y. So at x equals 1, we have y equals 0. That's just, again, that's e to the 0 equals 1, since it's all reversal. So the y is there. So if x is 1, y, y is 0. So this is going to go, lawn 1 is 0, yeah, 0 like that. So this equals to limit as T approaches infinity of lawn t squared over 2. And lawn T, well, that just goes to infinity. This just keeps going up and up and larger and larger. That's just our Y function. So this just goes to infinity, like that. Yes, so since this improper integral is divergent, the series itself is divergent, again, by the integral test. You just got to check if the, if it's decreasing, positive, and continuous. Then you've got to check if the integral diverges, if it does, then the series itself, the summation of lawn n over n from n equals 1 to infinity, is also divergent by the integral test. Otherwise, if it's convergent, then it will be convergent as well. So this improper integral, if it was reverse, it would have been convergent as well here.