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1. Draw a picture to show that

$ \displaystyle \sum_{n = 2}^{\infty} \frac {1}{n^{1.3}} < \int^{\infty}_1 \frac {1}{x^{1.3}} dx $

What can you conclude about the series?

converges

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Campbell University

Baylor University

University of Michigan - Ann Arbor

Boston College

to demonstrate the desired a comparison. We'LL just use a rough sketch Graff for one over X to the one point three power look something a little bit like this. Probably a little bit steeper, but the sake of this is not too important now. If we mark, this is one he's on. When we were talking about our syriza's, they're all going to do with one rectangles, starting with the height at, too. Since we're sort of equals two. That would be a rectangle. Something like this. You have another rectangle here and so on. However, the integral itself is the entire area underneath the curve, which starting at one, means we would include this area in addition to that of the rectangles. So we can see right away that free choice of Interval. There's a little bit, too gets missed bythe Siri's that the integral itself picks up so clearly. The area of the rectangles for Siri's is less in the area of the integral itself. Now, as far as what we can conclude about the serious based on this information, we can see if we actually evaluate that indefinitely. Nero Chrissy, if we take this integral Ah, using, of course. Ah limit. Since this is an improperly, necro will eventually see that this integral does in fact converge. Ah, you know, we do a little bit of magic and we get negative ten over three times, one over T to the one point three plus ten over three, which has t goes to infinity. This term becomes zero and we're left with just ten over three. Since the integral itself converges and thie, Siri's is less than that of the interview has a lower value than the integral. The Syria's itself must converge.