 Hello, and welcome to a screencast today about partial sums of an infinite series. Okay, so the nth partial sum of the series, a sub k, as k goes from 1 to infinity, is the finite sum of the first n terms. Okay, so let me emphasize here that this is going to be a finite sum. So that means it actually should hopefully have a number to it, and then of the first n terms only. Okay, so this is defined to be s of n is going to equal the sum as k goes from 1 to n of our a sub k. All right, the series we're going to look at here today is the sum as n goes from 1 to infinity of 1 over n minus 1 over n plus 1, and I want to calculate the first four partial sums, so s of 1, s of 2, s of 3, and s of 4. And then we'll see if we can see a pattern to them. All right, so following what we've got here for our notation, how we're defining these, so s of 1 is just going to be simply a sub 1. Okay, s of 2 is going to be a sub 1 plus a sub 2. s of 3 is going to be a sub 1 plus a sub 2 plus a sub 3. And then s of 4 is going to be a sub 1 plus a sub 2 plus a sub 3 plus a sub 4. Oopsie, and that's an a, not a 9. Sorry about that. Okay, so hopefully you can kind of see here there's a pattern to these as well, right? So like here for my s sub 3, I'm really just taking my value I got from s sub 2, and then I'm adding my new term a sub 3. Same thing for s sub 4. I'm really taking my value here from s sub 3, and I'm adding on my new term a sub 4. Okay, so that's kind of a, I wouldn't say a quick way to figure these out, but anyway it should hopefully help us a little bit. All right, so s of 1, so that's simply going to be my first term. So I'm just going to go ahead and take 1 and plug it into this particular formula then for my series. So that's going to give me 1 over 1 minus 1 over 2. So when I go to simplify that, I'm going to get a half, okay? All right, my s sub 2, so instead of me figuring out a 1, I already know what that is, right? That's going to be my 1 half, so that's my s sub 1, plus now what's my a sub 2? So my a sub 2 is going to be 1 half, so I'm going to plug 2 into this function now. So 1 half minus 1 third, which is going to end up giving me, let's see, so this is my a sub 2 term. Okay, so if I do 1 half plus 1 half, that gives me 1 minus 1 third, so that's going to give me 2 thirds. All right, following the same idea then to figure out what my s sub 3 is, I'm just going to go ahead and take my previous, my previous sum, so that was 2 thirds. So that's my s sub 2, and then now I'm going to add on to it, what's my term then for a sub 3? So if I plugged 3 into my function, that's going to give me 1 third minus 1 fourth. So I kind of am noticing a pattern here, so I add 2 thirds and 1 third and I get 1, right? I subtract off my 1 fourth, that's going to give me 3 fourths. Interesting. Okay, let's do our s sub 4 and let's see if this pattern continues. So again, I'm going to take my s sub 3 that I just figured out, so that's going to be my value of 3 fourths, plus I'm going to go ahead and then add on my term of a sub 4. So that's going to be 1 fourth minus 1 fifth. Again, I add 3 fourths and 1 fourth, I get 1, I subtract off 1 fifth, that gives me 4 fifths. Well, I'm seeing a pattern here for my partial sums. So if I were to generalize this, I could say that s sub n is, well, let's see. So s sub 1 gave me a half, s sub 2 gave me 2 thirds, s sub 3 gave me 3 fourths. So would you guys buy that this would give me the pattern of n over n plus 1? Interesting, okay. So we could actually say, and I know the problem didn't ask for this, but we could say then because now we can take a look at the limit of these partial sums and we can see if these actually converge. So let's do the limit as n approaches infinity of n over n plus 1. Okay, well my numerator is going to get really big, my denominator is going to get really big, but if we apply L'Hopital's to this, I'm going to get the limit as n approaches infinity of 1. So my limit actually gives me a value of 1, so that means that my partial sums will converge. Fascinating. Now, this will not always happen as it turns out this type of a series or sequence, depending on, you know, whichever one we're looking at here, but since we're summing things, this is a sequence or a series, sorry, this 1 over n minus 1 over n plus 1. This has a special name and it's called a telescoping series. Okay, just for future reference. All right, thank you for watching.