An awesome infinitely nested radical.

[Music] here i've got a nice limit which was on an international mathematics competition of the following name which i cannot pronounce without help so i won't try it's from the year 2003. and it was suggested by a good place to stop so our goal is to determine the limit as n goes to infinity of this crazy nested thing involving roots so let's look at the first four terms of the sequence to get a feel for what it looks like and then we'll parse out what this is so it starts with the square root of 1 and then the n equals 2 term will be the square root of 1 plus 2 and then we have the square root of one plus two times the square root of one plus three then we have the square root of one plus two times the square root of one plus three times the square root of one plus four and then so on and so forth so if you put all that together you'll end with n minus 1 times the square root of 1 plus n so here n minus 1 is 3 and n is 4 as you can see so generally when you see nested radicals the first maybe hint of what to try would be to find some recursion between the nth term and the n plus first term but that's actually pretty tricky right here because of how this becomes more deeply nested with these different coefficients so instead what we'll do is generalize this to like a larger family of sequences that we can compare among each other and then we'll find the limit of this sequence and so this is adapted from the official solutions so i thought i should probably say that so let's define the following sequence a m n so this is like a sequence with two indices so it'll look like this the square root of one plus m times the square root of one plus m plus one all times the square root of 1 plus all the way up to the n minus 1 times the square root of 1 plus n term so all that differs between this sequence and our sequence is the starting point for this nesting here we're starting at 2 and here we're starting at m so the nesting here is less deep if m is larger so if m is 5 this looks like 1 plus 5 times the square root of 1 plus 6 times all the way up to n whatever n is so let's point out real quick something that i've already said and that is our sequence is a two comma n so that's good to know then another maybe sequence that's important to look at as motivation for why we should be looking at this in the first place is the last one of these sequences which would be a n comma n and let's notice that a n comma n is just the square root of 1 plus n and that thing's fairly easy to deal with just because there's not much nesting going on at all so if somehow we can build a relationship between our sequence and this last sequence then we're probably good to go and we'll build that relationship through this definition of all of these links between them so like a2n maybe is linked to a3n which is linked to a4 and all the way up to a n n which is easy to deal with so now that we've kind of described the strategy and why we've adopted this strategy let's do maybe the obvious thing and let's square both sides of this definition so we can get rid of some of these square roots so maybe i'll say let's notice that a m n squared is equal to 1 plus m times the square root of a bunch of stuff but if we hone in on all of this stuff that's happening here we see that that is exactly am plus 1 n and i guess i should say this thing extends to include the square root so here we have am plus 1 comma n so that's good to see we're able to write a m n in terms of something with a larger m index which is good because the last n index is easy so if we can somehow push this higher and higher and higher then we'll be good to go but as you see we've got a square over here and we do not have a square over here so how could we maybe get rid of the square on this side well this is a bit tricky but it involves adding and subtracting the same thing from both sides of the equation to build a nice factorization and what we'll add and subtract is the following so i'll take a m comma n squared minus m plus 1 squared so i'll subtract m plus 1 squared from both sides of this equation so that'll be equal to 1 plus m a m plus 1 comma n minus m plus 1 squared like that this might seem like totally ridiculous but it does set up a factorization using a difference of squares here so we have a m comma n minus m minus 1 times a m comma n plus m plus 1 over here and then we can multiply this out and notice that we get m squared plus two m plus one since we have a subtraction there this one here will cancel this one here and then everything has an m in it so let's notice that that m can be factored out and we have m times a m plus one comma n minus m minus two again that's by factoring everything out okay i'm going to do one last very simple thing before we move everything to the top of the board and set up an absolute value here so this term is most definitely positive but we're not so sure about the other terms so i'll put absolute values around these and then we still have equality and then it's maybe a little bit easier to build an inequality okay so let's bring that to the top and then we'll work towards the end so on the last board we derived the following equation in terms of the generalization of this sequence it doesn't matter so much exactly what the terms of the sequence are anymore maybe we'll just notice that a two comma n was our sequence that we're trying to find the limit of and a n comma n was the square root of 1 plus n and then all of the rest of them satisfy the following equation so let's notice that this guy right here is most definitely not 0 so we can divide by that so that's going to give us the absolute value of a m n minus m minus 1 equals m divided by a m comma n plus m plus 1 times the absolute value of am plus n comma n minus m minus 2 and now we're headed towards a pretty nice position because we have described a m comma n in terms of am plus 1 comma n and if we can keep pushing that up we'll eventually get to this nice one right here we just have to worry about this kind of extra am comma n here but that's actually not too hard because we can make the following observation so that observation is that a m comma n is most definitely bigger than or equal to 1 which makes this whole denominator a m comma n plus m plus one bigger than or equal to m plus two but if this denominator is bigger than or equal to m plus two if we replace it with m plus two we push something larger in that direction so all of that will give us the following inequality so we'll have the absolute value of a m n minus m minus 1 is less than or equal to m over m plus two which is nice because we can maybe work with that times the absolute value of a m plus one comma n minus m minus two so now we've got a nice way to write lower terms in the m index in terms of higher terms in the m index so now let's move back to our sequence which let's recall that's a 2 comma n maybe just let's point out that we're moving back to our sequence so the absolute value of a 2 comma n minus 2 minus 1 that's the same thing as -3 so here m is equal to two will be less than or equal to two over four that can obviously be simplified to one half but let's just leave it like this and then the absolute value of a three comma n because we have a m plus 1 minus let's see 4. great so we've pushed it up 1. but now we can apply this again because the number we're subtracting is one larger than the index just like our situation here which builds this inequality so that'll give us this is less than or equal to two over four times three over five and then we have the absolute value of four comma n minus five and now we'll keep pushing that until we get n comma n let's maybe write one more step and then it'll be clear so we'll have two over four times three over 5 times 4 over 6 times the absolute value of a 5 comma n minus 6. so that's applying this one more time where m is equal to 4 in that case so now let's take that all the way to the top and by all the way to the top i mean to m equals n and that will give us 2 times 3 times 4 all the way up to n in the numerator and then let's notice that the denominator ends two further than the numerator so it'll be four all the way up to n plus two and then we'll be left with the absolute value of a n n minus n minus 1. and now let's bring that final inequality to the top and then we're ready to finish it off we just got done building the following inequality so a 2n minus 3 the absolute value of a2n minus 3 i should say where a 2n was a special case of our generalized sequence which was the one that we're interested in is less than or equal to this product of rational numbers 2 times 3 times 4 all the way up to n in the numerator 4 times 5 times 6 all the way up to n plus 2 in the denominator times the absolute value of a n n minus n minus 1. and now we're ready to apply the fact that we know a n n along with some nice simplification that happens with this rational number so let's notice that this 4 will cancel with this 4 this 5 will cancel with the 5 that comes out right after the 4. the 6 will cancel with the 6 that comes after the 5. then this n minus 1 will cancel with this thing that's 2 before the n plus one and then this n will cancel with the one that's one before that n plus one so notice that everything in the numerator except for two times three cancels and everything in the denominator except for n plus 1 times n plus 2 cancels furthermore we know what a n is so let's notice that this is equal to 6 over n plus 1 times n plus 2 times the absolute value of the square root of 1 plus n that's a n n and then minus n minus 1. but let's notice that the dominating term in the numerator is this n to the first power while the dominating term in the denominator is an n to the second power but that means as n approaches infinity this thing clearly approaches zero because the denominator is winning out if you will but if n approaches infinity causes this thing to go to zero that means that we have the limit as n goes to infinity of a to n in other words our sequence must be equal to three as its difference from three goes to zero and that's a good place to stop

Channel: Michael Penn

Views: 21,256

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: PVpoMvcWuCs

Length: 13min 27sec (807 seconds)

Published: Fri Dec 17 2021