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# Use the following steps to show that the sequence $t_n = 1 + \frac {1}{2} + \frac {1}{3} + \cdot \cdot \cdot + \frac {1}{n} - \ln n$has a limit. (The value of the limit is denoted by $\gamma$ and is called Euler's constant.)(a) Draw a picture like Figure 6 with $f(x) = 1/x$ and interpret $t_n$ as an area [or use (5)] to show that $t_n > 0$ for all $n.$(b) Interpret$t_n - t_{n + 1} = \left[ \ln \left(n + 1 \right) - \ln n \right] - \frac {1}{n + 1}$as a difference of areas to show that $\left\{ t_n \right\}$ is convergent.

## a) $\left[\sum_{k=1}^{n} \frac{1}{k}\right]-\ln n>0$$t_{n}>0$b) Both the areas as $n \rightarrow \infty$ are then positive, and since we found that the difference in areas approaches a positive value for large $n, t_{n}-t_{n+1}$ must be $> 0$ and therefore $\left\{t_{n}\right\}$ is a decreasing sequence.c) convergent

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##### Kristen K.

University of Michigan - Ann Arbor

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So we're given a sequence T end defined by the following s So here we have a partial some of the harmonic series, the first in terms. And then we subtract natural lug now for part A. We like to go ahead and show that this is always positive. Okay? And so here's how we can do this. So let's look at t in here. And I want to interpret this at least the first part here. Let's say, um blue as areas of rectangles. So the first one not put a rectangle of height one corresponding to the first term one and I'LL put it here and equals one to two. And this way the area of the rectangle is my first term second rectangle Hey, it will be a half for the second term and so on And then the very last one would start it and and plus one and that would have height one over. And however, let's also look at one over X And the reason I'm looking at one over X is because we'LL see in a second. Well, if we just integrate this So here, If I look at one over x and green Don't grass that this one will have area less than because it's always either on or below the boxes. So this number here is strictly less Then if you're looking at it in terms of areas so what that means is we can write the following. So just focusing on the apprentices we just showed that this is larger than and by simplifying the integral here as we just saw a moment ago, this is just Ellen and minus Ellen. And and there we go. As a consequence of our work here, we have t n larger than zero and that resolves part, eh? Now let's go on to part B here. So here we want to interpret the difference. Interpret this as a difference of areas to show that the sequence CN conversions. Okay, so first of all, let's just verified this expression here. Okay, so now after we subtract, we could cancel these and then watching out for the minus signs here will have Elland and plus one minus Ellen and minus one over and plus one. So now let's go ahead and follow the advice here. Let's interpret This has an area so firstly, she's focused on the first term. And so we mean, here's let's go ahead and replace this with the integral and sense and goes to infinity. We can integrate from one so infinity Ellen X plus one minus Ellen X. Similarly, if we interpret one over one plus and as an area, then the corresponding integral should be one to infinity, one over X plus one going on to the next page. Then the interpretation of the difference of the areas is interpreted as inner girl want to infinity, and now we just subtract. Now there will be some cancellation between these two intervals a little bit so instead of warring about limits of integration, But she's ignore those for a second, so ignore the limits of integration just for the time being so temporarily will bring these in at the end. We don't wanna have to use Lopez's house rule if in one integral, if we're gonna have to do the same exact argument in the second and the girl and the terms end up canceling out. So first, let's just focus on this indefinite integral hero here. We can use the integration by parts IBP and you would go ahead and just take you two be. And actually, in this case, let me actually take a step back before I integrate this. This one will be better if we write This is one lager them or even even better one plus one over X. Now go ahead and do integration My parts Here take your you just be the immigrant which forces Devi to just be DX. So if you do this integration by parts here, then you'll get the following and similarly the second integral including the sign here. This is just negative. And this is exactly why we wanted to ignore the limits of integration first. Now we could cancel those and then just take the limits and of integration to this guy so going on to the next page. So first, let me let me take a step back to just clear mentioned what I did here. So, first of all, we just showed this that the interpretation of the change in Ina girls, after doing in immigration, buy parts for the first Enter girl. We arrived at this term here. That was where the IBP came from and then the second in a girl and then we cancelled and now we just have However, we're instead of doing the indefinite girl We really should be looking at this in a girl here. So what we'LL do is right that infinity is a limit since we've already integrated. Now we just use the immigrant here, the answer here and then t is going or exes going from one to t. It's a plug in the tea and then if tea is one, you'd just get Ellen to there and then this is is in the limit. This is of the form infinity times Ellen of one which is zero. So we this is indeterminate, So we should prepare ourselves to use low Patel's rule. So what I'LL do here is right there Negative Ellen to out there and then I'll write this limit as ln over one or roti. Now this is of the form zero over zero as t goes to infinity, so use low Patel's rule here. They're a bit, um of the top one over one plus one over t use the chain rule with the power rule there and once again using the power rule down here. But luckily, these two terms cancel out from the chain rule and Steve goes to infinity. This will just goto one. So we just have one minus Ellen too, which is about point three oh seven. So going on to the next page, we conclude the difference in areas as n goes to infinity is positive and we have another remark here, Ellen and plus one over n is always positive, and it's increasing since that just the natural log is increasing. As Ex gets bigger, the natural larges gets bigger. If n gets bigger than this, fraction gets bigger and that ultimately means the log is getting bigger. Okay, so also we have both areas are positive as n goes to infinity because we have the same remark here for one over and plus one. This is positive but decreasing. And now we showed that the difference is approaching a positive number. And so what? This also really means is that's he and minus t n plus one is a must be larger than zero. And that means that this's a decreasing sequence because if you just push the Tien plus one to the other side, you would have this, and that's exactly what it means to be decreasing. And that resolves part B

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##### Kristen K.

University of Michigan - Ann Arbor

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