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Determine whether the series is convergent or divergent by expressing $ s_n $ as a telescoping sum (as in Examples 8). If it is convergent, find its sum.

$ \displaystyle \sum_{n = 4}^{\infty} \left( \frac {1}{\sqrt n} - \frac {1}{\sqrt {n + 1}} \right) $

$\frac{1}{2}$

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Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

let's determine whether the syriza's conversion or diversion by expressing s end the impartial son. So this s end. So remember, that's just in this case, that's just a four a five all the way up to a N. And I started it for because that's the starting point here and this is always are an So we write that as a telescoping some and then we'LL see if it's a conversion. If it is, conversion will find the some. Otherwise we could ignore this last line right here. So let's just rewrite this by first. Writing is a limit instead of looking at the whole sum. Well, just look at it s k and then like a goto infinity and this will be the whole song. So this is Lim Kay goes to infinity and SN is defined as the sum from four all the way up to ten salacious. Write that in the sigma notation. In this case, we're using s case over adding all the way up to Kay And then we have a k here, which is or in this case, a n. So we put one. Prentice is there for the summer. And then when we put another one out here because before we take the limit, this is our escape right here, this whole term in the parentheses. This is what we replaced NK with now using the instructions appear. Let's right. This is a telescoping some. So let's go out and rewrite this s k. So come down here. I have the limit. Kay goes to infinity. And then now let's just go ahead and expand the summation. So we plug in for first for end. That's our starting point. That would be my first term there when I plug in and equals four and then I will keep plugging in larger. And so the next one's five. So an equals five here. So this was any Poles for and equals five, six and so on. And I would keep going in this fashion and you could kind of see some cancelation. There already will come back to this in a second, And then I would keep adding, and eventually we'LL go all the way up to Kay. So that will be the very last term when I plug in K. But sometimes it helps to write. Maybe it's firmer to before that So let me write one more term here. So this is when you plug in and equals K down here and the one right before that is K minus one. So I'd have a one over K minus one, minus one over then I would want to do. Okay. Now, let's see if we could see the pattern here for the first term. One over. Radical four. There's nothing to cancel with it because all the other denominators are larger than before. Both for this negative one over route five. We could cancel it with this one. Negative. One of a radical six with cancel with the next one. And it looks like we could continue to do that even all the way until we get up to one of our okay, because that negative will cancel with the positive. So everything before in between the six and the K would have canceled out the one over roof or still there. And this term, at the very end, is still here, too. So let me go ahead and write this on the next patient's I'm running out of room, so we still have a limit out here, and we have these two terms On the inside of the limit, Lim Kay goes to infinity and then we had won over roof or or let's just rate is one half. However, when we take the limit here, the limit of one half is just a half. But as we take Kato infinity, the denominator goes to infinity. Let me write that this way. So the denominators going to infinity and the numerator is on ly one. So that will go to zero. And we got our final answer, which is one half. So the theories converges and we just found the sum of the Siri's, and that's our final answer.