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(a) Use the sum of the first 10 terms to estimate the sum of the series $ \sum_{n = 1}^{\infty} 1/n^2. $ How good is this estimate?

(b) Improve this estimate using (3) with $ n = 10. $

(c) Compare your estimate in part (b) with the exact value given in Exercise 34.

(d) Find a value of $ n $ that will ensure that the error in the approximation $ s \approx s_n $ is less than 0.001.

a) $1.54977, \quad$ error $\leq 0.1$

b) 1.6452222

c) 0.000288

d) $n>1000$

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Missouri State University

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Okay, So for part eight year, let's go ahead and use s ten to approximate or estimate the infinite some s And also we'LL also ask How good is the estimate? And so here by as ten of course we mean the end of the ten partial some. So the solution for part A So first will write out as ten. By definition, this is the sum of s just for the sum from one to ten. So in that case, just go ahead and write out the few terms here and add up a bunch of fractions. This would take quite a while here so one could go to the calculator, Miss round off here it's a one point five four nine seven six eight and that's that's an approximation to us. Now the answer, The second question, How good is the approximation? Well, if you look at exercise thirty four you'll see that they give the value for us for this sum here the infinite sum given by Oiler. So in our case, the error is the difference between the exact value in the approximation. So since we now have the exact value a plug that in and we'LL also play guitar approximation as ten, and in this case we're getting about point zero nine five two, which is less than one over ten. So it's not a bad approximation because it's less than one over ten, but we can do better by taking larger. And so it's not a bad estimation, considering we only use ten terms, but one over ten just might not be accurate up, depending on one's purposes. So let's go on to part B now for part B. This is world will actually go ahead and improve the estimate from party, eh? Using Inequalities three and plugging in and equals ten into three. So what's good and solve this so solution? So first, let's recall from party what we found. We found that s ten was about Okay, so now if you will get inequalities three it gives up or lower bounds through the exact value in terms of sn and these inner girls here. So notice the difference in the lower bounds of the intern girls ones and plus one the other one's the end. So here, because we're to use any equals ten, so that determines that. And if you recall from our Siri's and has won over and swear. And so that means FX should just be one of the explorer. So let's go ahead and plug all this in here. Okay? So then now we could actually just go ahead and integrate this we do have improper and liberals here, but these ones are not too bad, because when you plug an infinity, the expression become zero. So that would become s ten plus one over eleven, less than or equal to s less than or equal to as ten plus one over's him. So then now one would use the information from part, eh? And plug that in for the S tens here. So plug those in for as ten and then add the fraction and so going on to the next page. All right, this out. So there's s ten plus one over eleven and then s ten plus one over ten and then just add those together. And there we go now to find value for us here. So think of it this way. We have an interval here, a lower and upper bound. And all we know is that as you fall somewhere in between here. So instead of choosing either then points another approximation would just take the midpoint of these two the average and so here will take us to just be the average. Here. So is an approximation to s. And this becomes also we'd like to know the ear involved here, and so there's a few ways to do this. So this is our approximation for us and so sensitive in the middle, the air is just they have the length of the interval. And similarly, you can do this half if you like, but because we chose the mid point these two halves of the same length. So that's one way to write it. Or you could just take the entire honorable length and divided by two. So the entire in ruling it would be the right most end point minus the left most end point and divided by two. So this is Yeah, this is the length of the interval here. And if you go to the Congo later, you see that this is indeed less than the point zero zero five. So the ear is improved. Bye. Using in part B by using the midpoint s Now, let's go on to the next cage for party. This is where we'Ll compare the estimate in part B with the exact value that we mentioned in part A. But this is a coming for number thirty four. So this is in number exercise number thirty four. So solution. So from party, we had a approximation improved estimate for us. So let me write that down five to two to and we also saw what the ear bound was. So if you look at number thirty four as we mentioned, this gives the exact value dudes Oiler. So in this case, we see that the air given by using this value of s go to the calculator here zero zero zero to a and that's even that's also Weston the point zero zero five. So this shows that the answer we're getting from part B is getting very, very close to the exact answer here. Okay, so we have one more party here party. This will be the final part. This is where we'Ll find a value and you just need one value that'll ensure not the ear and the approximation is less than point zero zero one. Okay, so solution Well, we don't We don't want to use an equals ten anymore because that's not a good enough approximation. So instead, we're just we won't plug in a value for any it. But we'LL use this inequality for the air here. So for this is for the remainder when using in terms. So this will be one over X squared. That's R FX. All right, so it's pointing at that's remind us where the one over exports coming from. And we want this to be less than point zero one. This is what we want. And so we're gonna solve end to make sure we get what we want. So let's just go ahead and evaluate that in a girl there that'LL be a negative one over X from the power rule. And then if you plug in infinity, you get zero. If you plug in and you get one over end, simplify this and then we get an is bigger than one over point zero zero one, which is a thousand so any value of and larger than a thousand would work. So the smallest one that you could possible use is a thousand won any larger and would work, but here all they asked for is a value. So in that case, we could just use this value here and by what we just showed taking this value of and will ensure that the exact answer minus this value here, the one thousand and first partial. Some will indeed me less than point zero zero one, and so that resolves the problem.