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Numerade Educator



Problem 7 Medium Difficulty

Test the series for convergence or divergence.

$ \displaystyle \sum_{n = 2}^{\infty} \frac {1}{n \sqrt{\ln n}} $




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Video Transcript

No, this problem will use the integral test for the interval test. We want for these terms to eventually be strictly decreasing an absolute value to zero. So for that to happen, you'd want for this toe eventually, Just be increasing. So typically speaking, if you know that these things air going toe be going to zero, then you can guess that they're going to be going to zero an absolute value unless you have some trig functions involved. Because with Trig functions you can imagine there being some oscillations occurring. But generally speaking, it's a fairly good gas. If it looks like it's going toe zero, then it's probably going to be going to zero when you throw the absolute value signs around it as well. Like I said, the exception typically being with trig functions, you have to be a little bit more careful with that. Ah, but it is true if you want to be really rigorous and you should look at this, you could take the derivative of this and show that eventually the derivative is going to be positive. The derivative of this is positive for and large enough then that means that eventually this is just going to be decreasing to zero. Case of the integral test is applicable here. So the integral test says that this thing is going to be is going to have the same behavior. Is this integral? Okay, so this is ahh u substitution problem. So the U would be Ellen of N D'You would be won over and Tien So then the integral and consideration turns into Alan of to infinity. You to the minus one half. See you. So this lower bound here, we got that from looking at the lower bound over here. So over here are lower bound was too. And that correspondent to end. So when in is too plugin and has to hear and we get that you is natural log of two. So that's why when we're working with d ur Lord bound his natural log of two Get the upper bound here. We looked to see what's happening with up around here. So hear this corresponds toe in being infinity. Go on in his infinity, you is natural log of infinity, which is still infinity. And then we replaced the one over N d n a with d u. That's what this equation tells us we Khun d'Oh And then we had Then we're left with Do you want over square root of natural log of end, but natural log of in is just you sore left with one over the square root of you which is just you to the minus one half. Okay, so, of course it would be if you're working with D'You You want everything to be written in terms of you. So that's what these equations tell us to dio We know what to do with the d n. We know what to do with the natural lava vent. We we can use these equations to express everything in terms of you. And we can change the bounds by, you know, looking at the bounds appear and figuring out what the bounce correspond to buy this top equation. And then once we have this guy, this is just we could just use the power rule. We get one half, you do one half, and then we're evaluating this from natural log of the two to infinity. And this is certainly going to be something that is infinite. All right. If we plug in infinity here, we're going to get something that blows up. And then we're subtracting what happens when we plug in natural love too. And for you, So have infinity minus something finite. Certainly going to be infinite. Okay, So since this integral turns out to be infinite, the integral test tells us that this son there's also going to be infinite, which means that we get divergence.