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Use the sum of the first 10 terms to approximate the sum of the series

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

Use Exercise 46 to estimate the error.

$$s_{10} \approx 1.98828, \quad$ error $\leq 0.012$$

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

here, we'd like to use the first ten terms. So that will be the tenth partial, some as ten to approximate this infinite sum here from one to infinity. So now, as ten. The first, some of the first ten terms here. So we won't write all these up less used before we would answer this and a calculator. This is what we would end up answering here all the way up to N equals ten. And this would take too long here to write this out by hand. So going to a calculator, approximate this and that That will be one of our answers, the approximation of the infinite son. The question would be, and that's where the air will come in is how close is this approximation to the actual infinite sum? Because ten terms isn't that many. So it might lead you to think that the approximation is not a good one. A question mark there, and that's where the air will come in. So let's not the following that if we define our in to be the consecutive ratio of a N plus one over a n, where this is my end here and then divide by n. And if you take a limit of that as and goes to infinity, this will just become one and then one half. Cancel those two to the end. You just have one over two. Also, Oren is decreasing, decreasing sea ones. One way to show this. So if you look at our inn, we could simplify our previous are and appear before I canceled out that could have been written as and plus one over to end. Now, to show it decreases, you have to either show this or if you define f of X to be X plus one over two and this is usually preferred and works better. In many cases, the second one here show f Prime of X is negative If X is bigger than one, the's are both true here. And so I would recommend just writing this one out and then using the potion rule. So now the reason I'm checking these conditions here is because we want to use exercise number forty six and I'll go to the next page and write this out. So from number forty six, we checked the hypotheses that are needed on our end. So now we can say the following that the remainder after adding ten terms capital Orin is less than or equal to this term here. So this is all due to this previous problem here. So now just using n equals ten and plug it in and we see what we get that gives us our ten and then here will have eleven are eleven. So remember, definition of our end is a and plus one over and which so here we have our eleven, so that will be a twelve over a eleven, and this could be simplified is a fraction where you could just go to the calculator with this is well, and that's the air that we get. So in the previous page, we gave the approximate value. I'Ll recite that here, and the air was on ly this large and that resolves the problem