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The figure shows two circles $ C $ and $ D $ of radius 1 that touch at $ P. $ The line $ T $ is a common tangent line; $ C_1 $ is the circle that touches $ C, D, $ and $ T; C_2 $ is the circle that touches $ C, D, $ and $ C_1; C_3 $ is the circle that touches $ C, D, $ and $ C_2. $ This procedure can be continued indefinitely and produces an infinite sequence of circles $ \left \{C_n \right \}. $ Find an expression for the diameter of $ C_n $ and thus provide another geometric demonstration of Example 8.

$\sum_{n=1}^{\infty} a_{n}=\sum_{n=1}^{\infty} \frac{1}{n(n+1)}=1,$ which is what we wanted to prove.

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so for this problem will need to keep our eye on the figure that shone with the circles. So here, let's start off by using the fact that this triangle pr eh, is a right triangle. So if we use for data is their own, don't just tells us that a piece where plus p r squared is a r squared, and then so here D n is the diameter of the end circle. So hear this just tells us that this is one minus D one over two. That's a P, and then PR is just one. And then a r is one plus de one over two. Square that whole thing and then go ahead and solve this for the one to get D one equals one half. So then there let's go ahead and let's go on to the next page. So, using the fact that the same type of reasoning triangle p. R B, it's Rachel Angle. So once again by Petar Gris. This time we'LL have won minus so first that let me use the letters she's made. So this is he be square pr slayer B R Square Now, using r R D's this's the one plus de Tu over, too. That's this's all PB. But then PR, that's just once flair. And then br is one plus de tu over too square going and saw that for D, too. And you'LL end up with one over six. We could keep going on in this fashion, and similarly, you can go ahead and look at Triangle PRC. This is a right triangle, so let's do one more value of D. So this time, by path, a glorious you get P C Square plus P R's where equal C R squared and then this would become using the DEA notation de Juan plus d too d three over, too. That whole thing would be squared once where that PR is one and then on the right hand side, one plus de three over to quantity squared. And this will imply D three is one over twelve. So has gone to the next page. And let's just take observed the values that we've already found. So here it looks like if we had against the pattern one over this is to the right, that is one times two six we can write that is two times three, twelve, three times For and using this pattern, it looks like d n The diameter of the end circle should be one over end with an plus one. There look. So this is true and they can also be verified using mathematical induction. So here the next step will be to verify this fact using induction. So the first one would be the bass case. D one equals one over one times one plus one. And we know this is true because we're really computer D one using the geometry and we had one half. So the bass case for the induction is true. Let's go to the next step. So here we suppose d n is one over and and plus one And now we'LL Look at we want to show you that the formula is also true for d N plus one. So that's what we would like. So here let me take a step back. So here you wait would use Pythagorean theorem again so we'd have some triangle p r n Let this be the center of the n plus one circle C n plus one. So this triangle here would be another right triangle Oh, so we'd use potash arenas. We have been all along, so by degrees would have and our square PRC flared and r squared and are using the d notation and are we can write that using the same pattern that we've seen all along for the exact same reasons using the picture. So that's an R that we square that PR has been one along and then the right hand side following the usual pattern, which again comes from Cynthia pajama Tree that was given in the figure. Now I'm getting all about a room here, so I will need to go to the back to the next page. So this implies that's a sigma up. They're getting a little sloppy. That's the sun I equals Want. And dee, I am doing is just rewriting the rewriting the some so that we can see how to use the induction hypothesis. Oops, that's a two up there. So now we'LL just keep working with this. This is equivalent to one minus Now, how can we rewrite this? We would like to rewrite this sum so we know d I using night hypothesis one over, ay, ay, plus one. This is important stuff right here. We're replacing the D. I with this so far for the induction, we assumed that the formula is true. This is also known a strong induction or assuming that the formula is true for all the end for all the values of ten. So since this sum is firm ally from one's head and the hypothesis holds for these devalues, that's why we replace it with the fraction. But then we still add the other term at the very end outside of the sigma. And then we just rewrite everything else as it is now. Our next step would be to go ahead and compute this, but this is a telescoping some. So if you go ahead and evaluate that that sigma and again I should have had Princess, he's around that if you evaluate this signal, you'LL just get one minus one over and plus one And then now we're still have the n plus one over two squared plus one one plus we're almost there. Some of them may go on to the next page, So now we could cancel those ones and now we cancel the double negative to get a positive one over and plus one the square that add one right hand side Been the same the whole time. Not going to change it now and now we'LL just solve this for d N plus one soul a square, both sides and then foil this thing out. D n plus ones Quantity Square all over four plus one equals one plus d n plus one plus d n plus one swear over four and then solve this for d N plus one and we have d N plus one equals one over and plus one and plus two. So by mathematical induction, we've shown d n equals one over end and plus one for all and for all natural numbers. And so here we have noticed that so going to step back this gives another geometric demonstration. Yeah, of example seven, which is basically which says that the sum I equals one toe end of thie I equals one. So in other words, the sum of all the diameters equals one and that's our final answer