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Use the sum of the first 10 terms to approximate the sum S of the series_ (Round your answers to five decimal places:)n6 + 14.51822Estimate the error. (Use the Rema...

Question

Use the sum of the first 10 terms to approximate the sum S of the series_ (Round your answers to five decimal places:)n6 + 14.51822Estimate the error. (Use the Remainder Estimate for the Integral Test ) error < 0.2

Use the sum of the first 10 terms to approximate the sum S of the series_ (Round your answers to five decimal places:) n6 + 1 4.51822 Estimate the error. (Use the Remainder Estimate for the Integral Test ) error < 0.2



Answers

Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error. $$ \displaystyle\sum_{n = 1}^{\infty} \frac {1}{5 + n^5} $$

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

Let's use the some of the first ten terms to approximate the Siri's. So, in other words, this entire series is approximately as ten, and that's equal to the sum from one to ten five, minus in and then co sign swearing in. Now I've approximated this in the calculator already and Wolfram, so you could pause the video, go ahead and write on a few of those decimals. So that's our approximation to the infinite sum. Now this is where we'LL go ahead and actually see what the errors. So first, the best way to go about the year is the first. Think about how you would show this thing convergence. So looking at this is less than or equal to fight to the minus end, says co sign, squared, and is less than or equal to one. And for this series, this converges. You could use the fact that it's geometric. So, for example, for the here. So this is the noted by our ten, and this's bounded above by the air that you would get from using the integral test. So this is explained in more detail on page seven thirty, and then here. If we were to use the integral test. So here this is the integral from ten to infinity, one over five decks. So here you have to show, of course, that the function one over five X So we have to show that it's positive, continuous. These air both clear, and the last one's also clear that is decreasing. It's decreasing because every time X gets bigger, the denominator gets bigger, so the fraction it smaller go ahead and integrate that that's one over five to the ex over Alan off one over five tend to infinity. And so this is approximately point zero zero zero zero three more zeros and then a six. So let's a decimal. And then after the decimal, we have seven zeros, followed by six. So this is an upper bound for an ear, so that gives us an estimate for the year. And that's our final answer.

Let's use the some of the first ten terms to approximate the sum of the Siri's. So on the left over here, this is the sum of the Siri's, and that's approximately equal to us. Ten as ten. We know that's just the sum after using ten terms. So let's go to my next cabin, Wolfram Alpha, where, if you can see, I approximated the S ten the some of the first ten terms, so you could pause the screen. Write down a few decibels here for this approximation. So now that's are approximate. And now we want to estimate the ear as a result of using ten terms. So they're here is that most are ten. That here, this is the remainder. After using ten terms, that's a notation in the book. Now, before we go on in this direction, let's define affects to be one over X. Excellent fourth. So notice that if it's positive here, we're looking at X bigger than or equal to one because of the starting point is one and equals one. So if this positives on his ex is bigger than your equal to one, we could see that F is continuous because the numerator and denominator Robles continuous, and then we can see that f is decreasing. So just show this part. We need to see that the derivative is negative. So here is the question rule. Then swear that denominator. So the denominator is always positive, but we can see that the numerator is always negative. We can pull out the negativity and then we're less with X squared. Plus for X Cube. That's Herman. The parentheses is positive that the nominators positive. However, this negative is always a negative number, So this thing is less than zero. That means efforts also decreasing. So here we would to show convergence for the original Siri's would use the integral test. And so we're using the upper bound for the ear that's given by the inner will test. So here the upper bound would be the integral from ten to Infinity E one over X over X for the fourth and going head and evaluating this in a computer. This is approximately point zero zero zero three five nine and then three, six four so of repeated that point zero zero zero three five nine three six four. So the ear is no larger than this decimal over here. So that's our estimate in our final answer

Let's use the some of the first ten terms to approximate the Siri's. So here's the Siri's on the left that's approximately as ten. And then let's find let's approximate as ten. Actually, here's the exact answer, or someone Wolfram Alpha you could see on the left over here. The sum from one to ten. That's the sum that we want. Here's the exactly answer. If you want, it is a large fraction, too right out. But if you'LL settle for the that's more representation, then you Khun do point one nine eight. So that's our approximation to the actual son. And then I will find the year from using ten terms. So there is less than or equal to our ten. That's the remainder when using ten terms to approximate the song. Now for this theories here. If we have to show whether this thing converges, we could do a comparison, say, with one over three, then and then for this. Siri's here. This convergence is geometric, but if we've let's to know by FX dysfunction right here, then you can show that African defense. Let's see if it's positive. That's true continuous. This is true as well. Even differential and it's decreasing. It's clear to see that is decreasing. Every time X gets bigger on ly, the denominator gets larger, so the fraction is the hole gets smaller, so depressing. So here we can use the upper bound that you would get from the integral test. So this in apologies justified on Paige three seventy Excuse me, seventy. So this is just the integral from ten to infinity. So this is the upper bound for the air when using the integral test. So let's go and evaluate that integral. So that ends up being That's an infinity of there. So this we can find the exact answer by just plugging in the endpoints. Or we can just approximate this part of the calculator. So that's a decimal. The little needle here zero point zero zero zero zero two. So this is the upper bound for the air, and that's my final answer.


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