calculate-the-first-eight-terms-of-the-sequence-of-partial-sums-correct-to-four-decimal-places-doe-5

Question

Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent?
$$\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$

Kevin Shryock · Accepted Answer

welcome number on My Name is Kevin Chirac. Let&#x27;s take a look at the 1st 8 terms of an infinite Siri&#x27;s that is denoted as follows. It&#x27;ll be n equals one to infinity of one over n Cubed. And let&#x27;s look at the 1st 8 1st 8 terms in this, uh, the Siri&#x27;s. So now we&#x27;re looking for 1/1 cubed, plus one over two cubed, plus 1/3, cubed all the way up until we get to 1/8. Cute, but these are simply the terms inthe e sequence. What we&#x27;d like to do is look at the partial summation. So let&#x27;s now look at F sub one, which would be the partial summation up till the first term. And so that&#x27;s just something to be won over one Cubed, which is one let&#x27;s look at as to well as To is going to be the partial summation from the first term until the second term. So now we have one plus 1/8 that&#x27;s going to be 1.1 to 5 for s three. It would be up until and including the third term, So now we&#x27;ve got 1.125 plus 1 27th So let&#x27;s take a moment and pull out a calculators. And if we type into our calculators one 0.1 to 5 plus 1 27th and get a total of 1.16 to 035 decibels So one point 16 203 and that would be s three. We could continue this process all the way up until we get Teoh, including that eighth term, and we will be wanting to look at whether or not this sequence converges. Now. One thing that I will tell you is that this is going to be a sequence that converges, and the reason we can tell is because there&#x27;s going to be more and more of these decimal places that are going to begin toe walk. In other words, they&#x27;re not going to appear to grow even as we add more terms. So the number is going to approach something. But these individual digits are going to begin Teoh that the change will be so small that are larger digits will remain unaffected. So I hope

J Hardin · Answer

let&#x27;s calculate the first a terms of the sequence of partial sons. So four decimal places. So that&#x27;s asking for us one as to all the way. There s a will have to write all of these out. Okay, Now we know by definition, that s n is just a one. That&#x27;s the starting point all the way up to I am so as one is just a one. So this is just the term that you get when you plug in and equals one here. So they&#x27;re Wanda, put just one. And then, for example, as to that&#x27;s a one plus a two. So when we plug in one, we just have one. And then when you plug in and equals two minus a house so that&#x27;s point five. And for the remaining values, if we like, we could even come over here Instead of adding fractions, we could just use Wolfram Alpha. So here is You could see this is the right code because we&#x27;re able to compute these partial sums. We&#x27;ve already done the first two. Now, let me add the first reasons they want, plus a two plus a threes. You can see by the signal notation. This is the sun for history in this four at the very in lets me know that I just want four decimals. So here we have point six six six seven. So since we&#x27;re just rounding off here, this may not be exact. SEL. We should use approximation to be precise, he er now adding the first four terms increase that arena floor point six two five zero. Now going to five terms. We&#x27;re halfway there. Change that formula Five, but and compute that. And we have point six and then three threes now going to six terms. Change that five foot six and we have six three one nine now onto the seventh. We have six three, two one. And then finally, going to that eighth and final term will get point six three two one again. So that&#x27;s that might tell us something there. So that that answers the first part of this question. We found the first eight terms round that off to four decimals. Then there&#x27;s a second part of this question. Does it appear that the Siri&#x27;s is conversion or diversion? Well, just by looking at the terms here, it does look like the limit exists. So here I would say conversion. And it looks like the limit of SN is about point six three two. And if you wanted a formal explanation for why the Siri&#x27;s convergence, you could actually just use the alternating Siri&#x27;s test and check that all the conditions are satisfied, and that&#x27;s your final answer.

Kevin Shryock · Answer

My name is Kevin Shock. Let&#x27;s look at the infinite serious, in fact, the infinite alternating Siri&#x27;s. So we&#x27;ve got a serious starting at N equals one until Infinity and it&#x27;s denoted by negative one to the end, minus one time or divided by n factorial. And let&#x27;s just look at what the 1st 8 terms of this would be. We can take a moment and list this out. This would be first negative one to the one minus one power, which would be negative one to the zero power. Be careful not to drop those parentheses and then one factorial on the denominator. We could then add that Teoh negative one toothy Tu minus one powers that would be negative one to the first power over two factorial. And then we would have that similarly, for 34 we can go all the way up until we got to eight. So be negative one to the seventh power over eight factorial. And of course, this would continue on past, but we&#x27;re just gonna look at the first few turns. Now we do find out whether that this series converges by looking at the partial summation. So let&#x27;s start off with a sub one. That&#x27;s BR summation from the first term to itself Negative one. To this Europe, our would be a positive one, and over one factorial would just simply be one. So we would say that the first in this sequence of partial summations is one Let&#x27;s now look at as sub to as sub two is going to go along with the participation up till two. So this is now going to be the one from s of one. But now we&#x27;re going to be adding negative one to the first part of the negative 1/2 factorial be negative 1/2. So we&#x27;re adding this negative 1/2 and we could say that all of that comes together to be one. Now we&#x27;re going to be onto deaths of three. So s of three is going to be the 1/2 from sf two, and this time we&#x27;re going to be adding onto it 1/3 factory also be 16 without taking out our calculator too much. We can just say that that&#x27;s going to be 46 which is going to simplify down to 2/3. And if you continue to do this. So s a Before s a five of six, etcetera, etcetera, etcetera. And just keep adding on more and more terms all the way. Until we get to that eighth term not to jump ahead a little bit, the S sub eight is going to be zero point. Well, let me let me list about real quick. Actually, that&#x27;ll be helpful. So the S sub four is going to be 0.6251 as a five is going to be 0.6335 at sub six is going to be, 0.63 to 2 as sub seven is going to be, 0.6324 And notice what&#x27;s beginning to happen here. But we&#x27;ll finish up the full value here, so we&#x27;ll do s Subait as of eight is going to be 0.6324 So 0.6324 and notice what&#x27;s happening here. We&#x27;re locking in more and more decimals as we continue this through. So that&#x27;s 2/3 as we know, is just going to be 0.66 repeating. And so this is the first time we see this six. That&#x27;s going to continue with us through the rest of our answers, and we can see that the similar pattern begins to occur with this three, where it begins to lock itself in and then we get another one with the two that&#x27;s next to it. And then the four finally locks in a swell. So we can. It appears as if this does approach a limit, and we could say the limit that we believe it is approaching is approximately 0.6324 So we could write that as R L. And that&#x27;s what all of this we think equals. So it&#x27;s just about as far as we can tell. Just through this, this brute force processes. It&#x27;s just about 0.6324 All right, I hope that was

J Hardin · Answer

let&#x27;s find the first aid terms of the sequence of partial sums. So the sequence is denoted in your books by sn here. We&#x27;re starting at one, and we go up to infinity. So we&#x27;d like to find the first eight terms here. So that&#x27;s s one all the way to s eight. And then we&#x27;ll try to see if this series is conversion or diversion. So as one by definition, is just a one and this is our a m right here. So a one is just one over one to the fourth plus one squared one half. You can write that as 0.5, that will be our decimal. And then as to we can write, this is s one plus a two. In general, we can always write S n is equal to s n minus one plus an so s one we know that&#x27;s just one half from over here. And then we add in a two So a two and then we&#x27;ll go ahead and simplify this. So actually I&#x27;ve using Wolfram Alpha. I&#x27;ve actually computed all eight partial sums here, and these are the fraction. So this one half is what we had in part one. So continuing in this fashion, this is s one as to s three, 4567 and eight. However, and this answer, they want four decimal places. So all do here is just rewrite the decimals that I get from Wolfram. So, for example, for us to we go to the second row. Here we have 11 half 11/20. Excuse me. That&#x27;s what we would get after simplifying this summer. Fractions. Over here you have 11/20 and that is just 0.5500 That&#x27;s four decimals again, and we&#x27;ll just keep going all the way until we get to s eight. As three. This is as two plus a three. So s two plus one over three to the fourth plus three squared and you would go over to the Wolfram. We see that&#x27;s 10 1/1 80 and then let&#x27;s go ahead and round off their That&#x27;s approximately 0.5611 again, we&#x27;re going to four decimal places. So that&#x27;s why we stopped there. So that s three. Now, let&#x27;s go to s four in a different color as four would be s three, which we just found. Plus a four, which is one over for the fourth plus four squared. We would go ahead and simplify that. Or use Wolfram and using Wolfram 6913. 12,240 the denominator. And then just round that off using a calculator. 24 places. That&#x27;s what I get. 240.5648 Similarly, for s five. So that&#x27;s s four. Plus a five or s four plus 1/5 to the four plus five squared. Right. So here we can take a peek at the Wolfram. We&#x27;re on the fifth term here, so we do have this large numerator and denominator in row five. So that numerator 450,000 5. 69 that&#x27;s all on the top. And then in the denominator, 795,600. And this is S five here. And when we round this off using the calculator, this is approximately 50.5663 So we have five down, three more to go, so I&#x27;ll go on to the next page since I&#x27;m running out of room here. Uh huh. So this will be as six and then using all from this is our fraction. So here we have 16 million, 693,000. 153 up top in the new writer and the denominator. 29 million 4. 37 and then 200. And let&#x27;s go ahead and round that off. And if you run this off, this is Rs six, and this is approximately using a calculator. I get 60.5671 We have two more to go. Here&#x27;s S seven. That s six plus a seven. Right. So that&#x27;s the previous answer, Essex. Plus this. So here I am not rounding off. This s six that I&#x27;m using is not the decimal. This is the exact value that we got over here. And this. I&#x27;ve been doing this method the whole time. All of my s values are firm. This wolfram table here, these are all exact values and for the approximation of approximating each one individually. So for s seven. Go ahead and simplify this or use the wolfram. 818 million. So all of this is in our numerator In the denominator were there in the billions. There&#x27;s 44242 to 800. And once again use that calculator to round off their This is 5675 approximately. That&#x27;s going to four places. And for the last one s eight, using the same method we&#x27;ve been using all along. So use Wolf from here. We get three billion and some change in the numerator. A lot of change there and then in the denominator five billion and some change. And if we round this off using a calculator, that&#x27;s 0.5677 That&#x27;s one of four places again, and it appears that the sequence will converge. It looks like we&#x27;re converging to around 5.567 ish, maybe eight. Here, let&#x27;s we can also Well, you can&#x27;t see it from here because this is fraction form. Let me exit this. So let me go to the next page. So we had s one. As to all the way to s A and we started with 0.5 0.55 0.5611 0.5648 0.5663 0.5671 0.5675 and then 0.5677 So looking at these numbers and it appears that it will converge to this is just we&#x27;re not supposed to actually give the exact answer here. We were supposed to say whether it appears that it converges. I&#x27;m saying that it will, and more or less it will be about 0.5, say 568 And there&#x27;s our final answer.

Kevin Shryock · Answer

Welcome back to New Murad. My name is Kevin Chirac. Let&#x27;s take a look at the infinite Siri&#x27;s um, NC denoted by the summation from n equals one to infinity of one over the natural log of N Plus One. And we&#x27;re looking for whether or not this serious converges and the way that we can tell is just by simply listing a few of these terms out. So we&#x27;re going to generate this. In other words, we&#x27;re going Teoh, step each of these and values up along the counting numbers so we&#x27;ll start with an equal to one because we have that notated here. So be one plus one. Well, then add that to one over the natural log of two plus one, and we&#x27;ll continue this process until we were to get Teoh the eighth term one over the natural log of a plus one. We wouldn&#x27;t want to take a look at the partial summation, so the partial summations would be what happens when we add just the first, however many terms. So for the 1st 1 for example, be one divided by the natural log of to, and that would get us one point for four Teoh six. Looks like 95 will include that one for the second partial summation, we would include up till the second term. So now we would add an additional natural logger one over the natural log of three and they&#x27;ll get us a total of two point 35 to nine 34 And that would be our second. A partial summation. Now we could do this one more time for S three and s four. So I understand a calculate that&#x27;s really quickly. So now we&#x27;re just going Teoh, do it for the natural log of four and that&#x27;s going to get 3.7 for 28 Looks like to A to we could then do it for the fourth. And that would get us three point 69 561 I think we&#x27;re showing promise. I think we might have locked in this digit. Now I don&#x27;t know that we&#x27;re gonna grow much above three. Let&#x27;s see, we add it. Teoh. The natural, the reciprocal. The natural log of six? Nope. Nope. They blew right past it. So now we&#x27;re at the harsh estimation. Up till five would be 4.25 37 to 7. Partial the partial summation up to six and all I&#x27;m doing is just typing this in. My calculators were working through together, so I got 4.767 6 to 5. Looks like seven. Let&#x27;s do two more, so the next one is going to be plus one over. The natural log of eight gets 5.2485 to 4038 And I&#x27;m going to say the last one for you because what I want to point out with the last few minutes this video is that what we would be expecting to have happened is similar to what I thought was happening in the green here. Some of these digits would begin toe lock in, and they wouldn&#x27;t continue to step up. But that doesn&#x27;t seem to be the pattern. This four continues to grow onto the five and all of the other future digits air kind of following along with it. So this seems like it is continuing to grow, and it&#x27;s continuing to grow in such a way that it is not approaching a visible living. So I am going to claim that from the 1st 8 terms. This sequence does not. The Siri&#x27;s does not imply that it is convergent. So I would conclude that his divergent based on the 1st 8 terms of the partial summation Great. I hope this helped out if it did take him.

J Hardin · Answer

let&#x27;s calculate the first eight terms of the sequence correct of four decimal places. So for this instead of going by hand, I&#x27;LL just use the calculator. So here I am in Wolfram Alpha And if we look at the first some we only look as one just adding wants her We should have just won over the cube root of one, which is just one. If you want to go for decimals, you could write it that way. Now go to as to if we would write that out. But of course, here we want the approximation. So this time I will Sum of two terms and then on the outside is to make sure that we&#x27;re rounding off in this five is to give four decibels after the ones digit. So this is why I have five take it for digits to the right of the decimal. So one, seven, nine, three, seven and so on. And we&#x27;ll keep going all the way up to s ay So for as three this time we&#x27;LL add three terms So let&#x27;s come back to Wolfram Alpha four Hate seven one. So that&#x27;s two point four aid seven one goingto four decibels and we&#x27;LL keep going. We have five more. So now we&#x27;re on number four. This is the sum of the first four terms. So going to four decibels we have three and then point one one seven zero. Now we&#x27;LL go to five terms three and then seven zero one eight That seven zero one day after the three seven Oh one a. That&#x27;s what we wrote. Now let&#x27;s go to a six when I&#x27;m riding six terms This time we&#x27;re going upto four point two five to one. That&#x27;s for a six. That was four point two five two one. Now let&#x27;s go to seven terms. You know, we&#x27;re on our last two terms here. Four seven, seven, four nine That&#x27;s four seven seven four nine. And now we go to the last term by plugging in and equals for the upper bound and eight. And now we have five point two seven four nine. So that was five points. Who? Seven four. Nine. So all of these are our answers. For the first part of the question, we&#x27;ve calculated the first aid terms of the sequence of partial sums. That&#x27;s s one all the way up to as eight. Now, for the second part of the question, does it appear that the Siri&#x27;s is convergent or divergent? Well, we start off that one. Then we jumped all the way up almost by point eight. Then we increase almost by two. Then we increased. I&#x27;m sorry. I lost out of order here. So first, second. Then we jumped up by about point seven again point six, their point five, another point five. And then increased point five again. So I would say that this is that version and you could actually prove it&#x27;s diversion by the pee test here. Piers won over three, and that&#x27;s our final answer.

Problem

Calculate the first eight terms of the sequence o…

Problem 3 Easy Difficulty

Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent?
$$\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$

Answer

Convergent

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