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# (a) Show that the series $\sum_{n = 1}^{\infty} (\ln n)^2/n^2$ is convergent.(b) Find an upper bound for the error in the approximation $s \approx s_n.$(c) What is the smallest value of $n$ such that this upper bound is less that 0.05?(d) Find $s_n$ for this value of $n.$

## a) 2b) $R_{n} \leq \frac{\ln ^{2} n+2 \ln n+2}{n}$c) 1373d) 1.9392966

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okay for this problem in party. Let's start off by showing that the Siri's conversions. So for the solution here, let's go to the inner world test. So here's our a n So we look at f of X defined as such. Now show a N is decreasing. You can either go ahead and show definition of decreasing or we can just show f is decreasing. Now, this is a lot of cases, usually easier in this case. Go ahead and take a moment. Seize the quotient rule and when you simplify, you get this expression which is eventually negative here. If you take ex larger than eat, this will mean Ellen X is bigger than one. So that one minus ln x is negative. So if his eventually negative that I mean the derivatives Negative. So that means f is decreasing. Now we go ahead and use the integral test. So let's go ahead and write this proper. Integral is the limit. He goes to infinity. Now, if you look that this problem in the book it uses C s, which is computer algebra system, and that will come in handy right at this step because the integral is too difficult here. So using the computer algebra system. So going into any computer or calculator and just integrating this expression here we get the following. We had a negative out here and then Ellen squared X So by that we mean ln of Elena Vex all over X and then from one city. Now let's go ahead to the next page here because I'm running out of room. So I'll go ahead and just plug in the end points Now on the next page. Still that are negative there, Ellen squared T to Ellen T plus two. So we plugged in the team first, and then you would plug in one. Simplify this and now you can go in and take a limit that goes zero Ellen t over tea is of the form infinity over infinity so you could use low Patel's rule. And this approaches the derivative of ln over the derivative of Team one over tea which goes to zero as t goes to infinity. So we just showed is this whole fraction will go to zero so that the limit goes to two. This means that the integral convergence and so by the integral test that the Siri's also has to converge. That's how we're applying the integral tests and that resolves part, eh? Let's go on to the next page for part B here. So So for part B, we'd liketo find out we're bound for the ear in the approximation s is approximately yes, and so remember, as is the entire sum Where's sn is the partial some. So here recall what the ear is given by this is given by this is also known as the remainder. Yeah, which is given by the following. So recall the upper bound given in the section for the remainder when using in terms. So these are the same in here. So on our case, because we have an F and our problem here, it's replaced that with the definition of F that we gave in the previous pages and now write out this proper integral like we didn't party. So using the computer Ellsworth system just like we did in part a to integrate there are negative there, Ellen Squared X to Ellen X plus two Oliver X, And don't forget your end points here he and now we saw in part a that this expression here will go to zero in the limit. We use low Patel's rule to show that now this second term here doesn't depend on t So we'll just got in Simplify this and there's our were bound for the air So we have our end over here less than or equal to. And here's our upper bound. So that resolves Part V and we actually use this again soon going on to the next page here we want the smallest value of end such that the upper bound is less than point No. Five So solution So we'LL use party here. This is where we have our upper bound for our in Now we want the smallest and such that this is less than zero point five. So just go ahead and set this guy equal to we're less than a point of five And now this point, we could just go to any computer algebra system or set these equal to each other and see what you get for N. And since we're in is a natural number, we'LL have to round up here So in this case, the smallest and that we should get is thirteen seventy three after rounding up. So any of these ends will work because and is decreasing. So we just go ahead and take the minimum. This will be the smallest and that'LL guarantee that this is true. Finally, going onto party, we wantto find sn for this and in party that we just found. So this is the end equals thirteen seventy three from party. So now we just go to the partial son. By definition, you go from your starting point all the way up to this new value, and then you just take the sum of the Siri's. Now that's too much for us to add up by hand. So there's many you could perhaps right up your own code. If you's Mathematica or Wolfram, you can answer this in as some. It's a capital s with Brackett there ln of end, squared over and squared. And now we want to give the index. So we're summing over end. We're starting at one ending at thirteen. Seventy three. Close the braces, close the bracket, and neither case when you answered this in and round off, we get this and that's our answer for party, and that results upon him

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