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2) Consider the infinite seriesCn=ln: Get an estimate for the sum of this series with error < 0.01 using as few terms as possible following the methods used in c...

Question

2) Consider the infinite seriesCn=ln: Get an estimate for the sum of this series with error < 0.01 using as few terms as possible following the methods used in class and in your text book: You will need to determine the number of terms needed (show all work), then use it to get_ a correct estimate: In the space below; give the estimate with 2 places after the decimal point:

2) Consider the infinite series Cn=ln: Get an estimate for the sum of this series with error < 0.01 using as few terms as possible following the methods used in class and in your text book: You will need to determine the number of terms needed (show all work), then use it to get_ a correct estimate: In the space below; give the estimate with 2 places after the decimal point:



Answers

In Problems 19-26, approximate the sum of each series using the first three terms and find an upper estimate to the error in using this approximation. $$ \sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k^{2}} $$

Okay, so our case starts here at zero and we are going to find the sum using the first three terms of 01 and two and then we're going to find the upper estimate to our area using that approximation. So the place is zero and negative. One to the zero power is 10 factorial is one and one half to the zero power is one. So our first term is one then for putting a one in, we will now be negative and one factorial is one and then one half to the first power is a half. Okay, so now we're putting in a two, we go back to being positive two factorial is just two and then to the second power, so that is going to be a 1/8 when we some all those together, that's 5/8. Or we can also think of that in our decimal as 0.625. Okay, now our error is about our next term. So the next term we put in a 012 and now we're putting a three in. So we have that 1/3 factorial, which is 1/6 and then that half to the third power. So that's an eighth. So that will be a 1/48. And so we can say that our error will be less than or equal to 0.0208, three.

We are going to approximate the some of our series using the first three terms and then find the upper estimate to the error in using this approximation. So kate starts at zero. So when I put a zero in negative, one to the zero power is just one and then 1/2 times zero plus one is one. And then one third to the first power is actually one third. So we will start with a one third. Now our next one putting one in that is going to be negative R two K plus ones. In both cases it will be a three. So we have a one third and then we have a one third to the third power. It's really one third to the fourth power. So that is a one over. I will say that in the next one that's gonna be 1/81. Okay, so this guy right here, if I put now let's see 01. So I'm putting a two in. So I will have a 1/5 and then also one third to the fifth power. So to consider what these numbers are, that's a one third minus 1/81 plus one over 1215. When we add those all together again, 0.3218. So our error is about our next term. So this time we are going to be placing the value of four in. So that's going to get us actually 012, actually replacing three in how to recount those. So we're putting the value of three in. So two times three plus one is 1/7. So we're going to get a 1/7 times one third to the seventh. So throw that into my calculator and I find that my error is actually less than or equal to 0.000065.

Okay we are going to approximate are some by using the first three terms and then find the upper estimate to our error. So with this notice that case starts at zero. So negative one to the zero power is one and also zero factorial is one. So we start with a value of one then the next one is negative. So when we put a one in um we just are going to have that negative one because 1/1 factorial and then putting a two in we're gonna be back to plus and then two factorial is just two times one. So that's just a one half. So it ends up our value is one half when we approximate that some with only three terms. And now the error. The error is always going to be in our um alternating is always going to be the very next term. So that's going to be putting a three in. And so it would normally be negative. But remember we want that top we want the upper um error so we make it positive. It was we can say 1/3 factorial and that's just three times too. So that will be six, So 1 6th is 0.16666 continuous.

We're going to approximate the some of our series using the first three terms and then find the upper estimate to the error. And using this approximation. So first we're going to be putting a one in, so negative one to the second, power is going to be positive And then I have a 1/1 to the power one. So everything ends up being one. Now putting it to in will make this negative and I have 1/2 squared which is 1/4. And then we go back to being positive. And when we put our three in we have a 1/3 to the third, which is 1/27. We had these all together, we get 0.1787. As our approximation of our series. Our error is going to be less than or equal to that very next term. And so consider the next term that we are putting in. We put in 12 and three. So we're ready to put K equals four in. And so our error is going to be less than one, divided by four to the fourth. And so our error will be less than or equal to 0.003, nine.


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