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Current time:0:00Total duration:12:31

Video transcript

so what we're going to attempt to do is evaluate this sum right over here this evaluate what this series is negative 2 over n plus 1 times n plus 2 starting at N equals 2 all the way to infinity and if we wanted to see what this looks like when it starts at N equals 2 so when N equals 2 this is negative 2 over 2 plus 1 which is 3 times 2 plus 2 which is 4 then when n is equal to 3 this is negative 2 over 3 plus 1 which is 4 times 3 plus 2 which is 5 and it just keeps going like that negative 2 over 5 times 6 and just keeps going on and on and on and now it looks pretty clear that each successive term is getting smaller and it's getting smaller reasonably fast so it's reasonable to assume that this might be a finite that you're even though you have an infinite number of terms it actually might give you a finite value but it doesn't jump out at me at least the way that I've looked at it right now as to what this sum would be or how to actually figure out that sum so what I want you to do now is pause this video and I'm going to give you a hint about how to think about this try to dig up your memories a partial fraction expansion or partial fraction decomposition to turn this expression into the sum of two fractions and that might help us think about what this sum is so I'm assuming you've given a go at it so let's try to manipulate this thing let's see if we can rewrite this as a sum of two fractions so this is negative two over and we can I'm gonna do this in two different colors n plus 1 times n plus 2 times n plus 2 and we remember from our partial fraction expansion that we can rewrite this as the sum of two fractions as a over N plus 1 plus B plus B over n plus 2 and why is this reasonable well if you're adding two fractions you want to find a common denominator which would be a multiple of the two denominators this is clearly a multiple of both of these denominators and we learned in partial fraction expansion that whatever we have up here especially because the degree here is lower than the degree down here what ever we have up here is going to be a degree lower than what we have here so this is a first degree term in terms of n so these are going to be constant terms up here so let's figure out what a and B are so if we if we perform the addition well let's just write rewrite both of these with the same common denominator so let's rewrite a over n plus 1 but let's multiply the numerator and denominator by n plus 2 so we're multiply the numerator by n plus 2 and the denominator by n plus 2 I haven't changed the value of this first fraction similarly similarly let's do the same thing with B over n plus 2 multiply the numerator and the denominator by n plus 1 so n plus 1 over n plus one once again haven't changed the value of this fraction but by doing this I now have a common denominator and I can add so this is going to be equal to this is going to be equal to n plus 1 times n plus 2 is our denominator and plus 1 times n plus 2 is our denominator and then our numerator our numerator let me expand it out our numerator this is going to be if I distribute the a it is a n plus 2 a so let me write that a n plus 2 a and then let's distribute this B plus BN plus B plus B n plus B now what I want to do is I'm going to rewrite this so I have all of the N terms so for example so for example a n plus BN I could factor an N out and I could rewrite that as I could rewrite that as a plus B times n those two terms right over there and then these two terms the two a plus B I could just write like this plus two a plus B and of course all of that is over all of that is over this whole thing is over n plus 1 times n plus 2 times n plus 2 so how do we solve for a and B well the realization is that this this thing must be equal to negative to these two things must be equal to each other remember where we're making the claim that this which is the same thing as this is equal to this that's the whole reason why we started doing this so we're making the claim that these two things are equivalent we're making this claim so everything in the numerator must be equal to negative two so how do we do that we're just it looks like we have two two unknowns here how do we you know to solve to figure out two unknowns we normally need two equations well the realization here is look we have an N term on the left hand side here we have no n term here so you literally could view this instead of just viewing this as negative two you could view this as negative two plus zero n plus zero times n that's not on that this is that's zero let me write it this way zero times zero times n so when you look at it this way it's clear that a plus B is the coefficient on n that must be equal to zero a plus B must be equal to zero and this is kind of bread-and-butter partial Frank Frank fraction expansion we have under videos on that if you need to review that and that the constant part to a plus B to a plus B is equal to negative two is equal to negative two and so now we have two equations in two unknowns and we could solve it a bunch of different ways but one interesting way is let's multiply the top equation by negative one so then this becomes negative a minus B is equal to well negative one times zero is still zero now we can add these two things together and we are left with 2a minus a is a plus B minus B well those cancel out is equal to a is equal to negative two and if a is equal to negative two a plus B is zero B must be equal to 2 B must be equal to two negative two plus two is equal to zero I just took I solved for a then I substituted it back up here so now we can rewrite this all of this right over here we can rewrite it as the sum and actually let me do a little bit in today let me just write it as a as a finite as a finite sum as opposed to an infinite sum and then we just take the limit as we go to infinity so let me rewrite it like this so this is the sum from N equals two instead of to infinity I'll just say to capital to capital and then later we could take this the limit as this goes to infinity of well instead of writing this I can write this right over here so a is negative two so it's negative 2 over n plus 1 and then B is 2 plus B over n plus 2 so once again I just express this as a finite sum later we could take the limit as capital n approaches infinity to figure out what this thing is let's start and B let me not write B any more we now know that B is 2 over n plus 2 now how does this how does this actually go about helping us well let's do what we did up here let's actually write out what this is going to be equal to this is going to be equal to when n is 2 when n is 2 this is negative 2 over 3 so it's negative 2/3 plus 2 over 4 plus 2 over 4 so that's N equals let me do it down here so I'm about to run out of real estate that is when n is equal to 2 now what about when n is equal to 3 when n is equal to 3 n is equal to 3 this is going to be this is going to be negative 2 over 4 negative 2 over 4 plus 2 over 5 plus 2 over 5 what about when n is equal to 4 I think you might see a pattern that's starting to form let's do one more when n is equal to 4 when n is equal to 4 when n is equal to 4 well then this is going to be negative 2 over 5 negative 2 let me do that same blue color negative two-fifths negative two-fifths plus 2 6 plus 2 6 and we're just going to keep going we're just going to keep going let me scroll down and get some space we're going to keep going all the way until the nth term so we're going to keep going all the way to the nth term so let me just so plus dot dot dot plus our capital nth term which is going to be negative 2 negative 2 over capital n plus 1 plus 2 plus 2 over capital n plus 2 so I think you might see the pattern here notice notice from our first when N equals 2 we got the 2 for its but then on N equals 3 to the negative 2/4 that cancels with that when N equals 3 add 2/5 then that cancels when N equals 4 the negative 2/5 so the second term cancels with the second the second part I guess when for each n for each index cancels out with the first part for the next index and so that's just going to keep happening all the way until and all the way until capital n until n is equal to capital N and so this is going to cancel out with the 1 right before it and all we're going to be left with all we're going to be left with is is this term and this term right over here right over here so let's rewrite that so we get some more space here this thing can be rewritten as so the sum the sum from lowercase n equals 2 to capital n of negative 2 over n plus 1 plus plus 2 over n plus 2 is equal to all everything else in the middle cancelled out we're just left with negative 2 thirds plus plus 2 over capital n plus 2 so this was a huge simplification right over here and remember our original sum that we wanted to to calculate that just has the limit as capital n goes to infinity so let's just take the limit as Capital n goes to infinity so let me write it this way though actually let me write it this way the limit so we can write it this way the limit as capital n approaches infinity is going to be equal to the limit as capital n approaches infinity of well we just figured out what this is this is negative 2/3 negative 2/3 plus plus 2 over capital n plus 2 well as n goes to infinity this negative 2/3 doesn't get impacted at all this term right over here to over an ever larger number over an infinitely large number well that's going to go to 0 and we're going to be left with negative negative 2/3 and we're done we were able to figure out the sum of this infinite series and this type of series so this thing right over here is equal to negative 2/3 and this type of series is called a telescoping series telescoping I should say this is a telescoping series telescoping series and the TEL a telescoping series is a general term so when if you were to take its partial sums it has this pattern right over here where in each term you're starting to cancel things out so what you're left with is just a kind of a fixed number of terms at the end but either way this was a pretty a little bit hairy but was a pretty satisfying problem