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# Graph the curves $y = x^n, 0 \le x \le 1,$ for $n = 0, 1, 2, 3, 4, . . .$ on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 8, that$\displaystyle \sum_{n = 1}^{\infty} \frac {1}{n(n + 1)} = 1$

## $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=1$

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Let's graph the curves. Why equals X at the end on the same screen where X is only going from zero one? So let's do an equal zero. So why equals X zero equals one? This is the first curve. It's just a horizontal line. Y equals one. Then you could try another number. Let's try and equals one. So you have y equals X to the one equals X. That's just this line. Some of the origin with Slope one more or less. Let's go for another ex cleared. This is our parabola and we could plot some points here to see that the proble actually lies underneath. So the common theme here is every time we draw new graph, it's actually going to be below the previous graft. And the reason for that is that when you take a number between zero and one, the higher the power you multiplied, the more times you are multiplying a small number that's gonna make the overall number the resulting number much smaller. So I'll do one more Here, let me go back to read y equals X cubed Then here If we were a plot this one you get something that's even lower than the blue. And we'LL just continue to have pictures that looked like this. So that's the grafts on the common screen. That's the first part of the question. Now let's find the areas between successive curves. So what this means is this triangle right here. This would be the area between the first two curves, so that would be a lot area one here between the green and the blue. That would be the area between the second and third curves and so on. So let's go ahead and find that, and by doing this will give a geometric illustration of this fact here. Infinite Syriza's people tto one. So we'LL try to explain why this equation is true by adding up all of these areas in the figure. So let's go ahead and well, we know that if we add so let's make an observation here. Let's just write it this way, since I've called a one plus a two plus a three plus that that what should this If I keep doing this forever, what's the sun going to go to? And I claim it will be one and the reason is if we continue to just add in these graphs. And if we just find all the areas A one a two, a three and so on were eventually going to get the area of the entire Tze Square. And that's just one times one bases, one hiders one. So the area, this whole area is just one. On the other hand, I can I know that the area between two successive curves top minus bottom. This is equal to a so really here I should do for a n. Let me take a step back here. So this should be a and plus one. So it looks like here I should start and to be zero. So if I had of all the A's and equal zero to infinity a a and plus one, So this is the quantity that we want. So that's equal to the sum from zero to infinity of exit E end minus exit A and plus one. And then I have to integrate this zero one. All I'm doing here is just using the definition of A to replace a with the stigmatization. No, let me bring this to the other side over here. Since I'm running out of room This we can write as the sum from zero to infinity. Now let's use our immigration soap our rule there power rule again. Now plug in those end point zero one So an equal zero to infinity and plug in the one in this era and then just gone and simplify this fraction. You're going to get a common denominator in there. So I'm getting one over and plus one and plus two. Or really, you could have also telescopes. I should've mentioned that you could telescope here if you want to. You get the same answer. But I hear the reason I woulda telescope is because we have a specific goal in mind and we would like to see this signal notation eventually arise in our work. That way we could justify the result. And so here, actually this we can do that there step. So I have the Sigmund that I want right here. I claim this were basically finished, but to make it more obvious if you notice our denominators different than theirs. But the reason that's okay is because they're starting at a different and then we are so here. Let's go ahead and use a different letter. Let's replace that end with Kay so n equals zero. This means Capel's want so I can rewrite this using a new sake my here. So let's use K equals one to infinity. And then I have won over and plus one that's just k and then n plus. Ooh, this is K plus one and this is the same as the sun that was already given. And we know that this must equal one because this is equal to the sum of all the A's. And as we already pointed out, the sum of all the A's equals one because that's the area of the entire square. So this argument, the counting the areas between the success of curves gives a geometric explanation for why this infinite sum is equal to one. And that's your final answer.

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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Join Bootcamp